This problem is same as the previous one, but has larger constraints.

Shiro's just moved to the new house. She wants to invite all friends of her to the house so they can play monopoly. However, her house is too small, so she can only invite one friend at a time.

For each of the n days since the day Shiro moved to the new house, there will be exactly one cat coming to the Shiro's house. The cat coming in the i-th day has a ribbon with color u_i. Shiro wants to know the largest number x, such that if we consider the streak of the first x days, it is possible to remove exactly one day from this streak so that every ribbon color that has appeared among the remaining x - 1 will have the same number of occurrences.

For example, consider the following sequence of u_i: [2, 2, 1, 1, 5, 4, 4, 5]. Then x = 7 makes a streak, since if we remove the leftmost u_i = 5, each ribbon color will appear exactly twice in the prefix of x - 1 days. Note that x = 8 doesn't form a streak, since you must remove exactly one day. 

Since Shiro is just a cat, she is not very good at counting and needs your help finding the longest streak.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the total number of days.

The second line contains n integers u_1, u_2, …, u_n (1 ≤ u_i ≤ 10^5) — the colors of the ribbons the cats wear. 

Output

Print a single integer x — the largest possible streak of days.

Examples

Input


13
1 1 1 2 2 2 3 3 3 4 4 4 5


Output


13

Input


5
10 100 20 200 1


Output


5

Input


1
100000


Output


1

Input


7
3 2 1 1 4 5 1


Output


6

Input


6
1 1 1 2 2 2


Output


5

Note

In the first example, we can choose the longest streak of 13 days, since upon removing the last day out of the streak, all of the remaining colors 1, 2, 3, and 4 will have the same number of occurrences of 3. Note that the streak can also be 10 days (by removing the 10-th day from this streak) but we are interested in the longest streak.

In the fourth example, if we take the streak of the first 6 days, we can remove the third day from this streak then all of the remaining colors 1, 2, 3, 4 and 5 will occur exactly once.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class A {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		long first, second;
		
		public Pair(long first, long second) 
		{
			this.first = first;
			this.second = second;
		}
		
		public int compareTo(Pair ob) 
		{
			return (int)(ob.second - second);
		}
	}
	
	static class Tuple implements Comparable<Tuple>{
		long first, second,third;
		public Tuple(long first, long second, long third)
		{
			this.first = first;
			this.second = second;
			this.third = third;
		}
		public int compareTo(Tuple o)
		{
			return (int)(o.third - this.third);
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static boolean isPrime(long n) 
	{
		if(n == 1)
		{
			return false;
		}
		for(int i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
}
	
	public static List<Integer> SieveList(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		ArrayList<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static int lcm(int a, int b)
	    {
	        return (a / gcd(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		//Count the number of coprime's upto N
		public static long phi(long n)  //euler totient/phi function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }	
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(long n, long k) 
	    {
			long ans = 1;
			for(long i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		public static void Sort(long[] a) 
		{
			List<Long> l = new ArrayList<>();
			for (long i : a) l.add(i);
			Collections.sort(l);
//			Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a)
		{
			List<Character> l = new ArrayList<>();
			for (char i : a) l.add(i);
			Collections.sort(l);
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   /*
		    * 
		    * >=         <=
		   0 1 2 3 4 5 6 7  
		   5 5 5 6 6 6 7 7
		   lower_bound for 6 at index 3 (>=)
		   upper_bound for 6 at index 6(To get six reduce by one) (<=)
		   */
		
		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1, r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}

	
			
	public static void main(String[] args) throws Exception 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int N = (int)(2e6);
		int[] f = new int[N];
		int[] ff = new int[N];
		
		long ans = 0; 
		for(int i = 1;i<=n;i++)
		{
			int x = sc.nextInt();
			f[x]++;
			ff[f[x]]++;
			if(f[x] * ff[f[x]] == i && i != n)
			{
				ans = (i+1);
			}
			else if(f[x] * ff[f[x]] == i-1)
			{
				ans = (i);
			}
		}
		
		fout.println((ans == 0) ? 1 : ans);
		
		fout.close();
	}
}
