You are given two strings a and b consisting of lowercase English letters, both of length n. The characters of both strings have indices from 1 to n, inclusive. 

You are allowed to do the following changes: 

  * Choose any index i (1 ≤ i ≤ n) and swap characters a_i and b_i; 
  * Choose any index i (1 ≤ i ≤ n) and swap characters a_i and a_{n - i + 1}; 
  * Choose any index i (1 ≤ i ≤ n) and swap characters b_i and b_{n - i + 1}. 



Note that if n is odd, you are formally allowed to swap a_{⌈n/2⌉} with a_{⌈n/2⌉} (and the same with the string b) but this move is useless. Also you can swap two equal characters but this operation is useless as well.

You have to make these strings equal by applying any number of changes described above, in any order. But it is obvious that it may be impossible to make two strings equal by these swaps.

In one preprocess move you can replace a character in a with another character. In other words, in a single preprocess move you can choose any index i (1 ≤ i ≤ n), any character c and set a_i := c.

Your task is to find the minimum number of preprocess moves to apply in such a way that after them you can make strings a and b equal by applying some number of changes described in the list above.

Note that the number of changes you make after the preprocess moves does not matter. Also note that you cannot apply preprocess moves to the string b or make any preprocess moves after the first change is made.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the length of strings a and b.

The second line contains the string a consisting of exactly n lowercase English letters.

The third line contains the string b consisting of exactly n lowercase English letters.

Output

Print a single integer — the minimum number of preprocess moves to apply before changes, so that it is possible to make the string a equal to string b with a sequence of changes from the list above.

Examples

Input

7
abacaba
bacabaa


Output

4


Input

5
zcabd
dbacz


Output

0

Note

In the first example preprocess moves are as follows: a_1 := 'b', a_3 := 'c', a_4 := 'a' and a_5:='b'. Afterwards, a = "bbcabba". Then we can obtain equal strings by the following sequence of changes: swap(a_2, b_2) and swap(a_2, a_6). There is no way to use fewer than 4 preprocess moves before a sequence of changes to make string equal, so the answer in this example is 4.

In the second example no preprocess moves are required. We can use the following sequence of changes to make a and b equal: swap(b_1, b_5), swap(a_2, a_4).
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;
public class A {
 
	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>
	{
		int first, second;
		public Pair(int aa, int bb)
		{
			first = aa; second = bb;
		}
		public int compareTo(Pair p)
		{
			if(first == p.first) return (int)(second - p.second);
			return (int)(first - p.first);
		}
	}

	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(int 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);
		
		List<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;
	    }
	
		public static int phi(int n)  //euler totient function
	    { 
	        int result = 1; 
	        for (int i = 2; i < n; i++) 
	            if (gcd(i, n) == 1) 
	                result++; 
	        return result; 
	    }
	
	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 nCr(long n, long r)
	    {
	    	long ans = 1;
	    	for(long i = 1,j = n;i<=r;i++,j--)
	    	{
	    		ans = ans * j / i;
	    	}
	    	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 mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		
		   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;
			    int 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 IOException 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		String a = sc.next();
		String b = sc.next();
		
		int res = 0;
		 if(n % 2 != 0) {
	            int pos = (n / 2);
	            if(a.charAt(pos) != b.charAt(pos)) res++;
	        }
	 
	        for(int i=0;i<n/2;i++) {
	            if(a.charAt(i) == b.charAt(i) && a.charAt(n - 1 - i) == b.charAt(n - 1 - i)) {
	                continue;
	            }
	 
	            if(a.charAt(i) == b.charAt(n - 1 - i) && a.charAt(n - 1 - i) == b.charAt(i)) {
	                continue;
	            }
	 
	            if(a.charAt(i) == a.charAt(n - 1 - i) && b.charAt(i) == b.charAt(n - 1 - i)) {
	                continue;
	            }
	 
	            if(a.charAt(i) != b.charAt(i) && a.charAt(i) != b.charAt(n - 1 - i) &&
	                    a.charAt(n - 1 - i) != b.charAt(n - 1 - i) && a.charAt(n - 1 - i) != b.charAt(i)) {
	 
	                if(b.charAt(i) == b.charAt(n - 1 - i)) res++;
	                else res += 2;
	 
	                continue;
	            }
	 
	            res++;
	        }
	 
	        fout.println(res);
		fout.close();
	}
}