Eshag has an array a consisting of n integers.

Eshag can perform the following operation any number of times: choose some subsequence of a and delete every element from it which is strictly larger than AVG, where AVG is the average of the numbers in the chosen subsequence.

For example, if a = [1 , 4 , 3 , 2 , 4] and Eshag applies the operation to the subsequence containing a_1, a_2, a_4 and a_5, then he will delete those of these 4 elements which are larger than (a_1+a_2+a_4+a_5)/(4) = 11/4, so after the operation, the array a will become a = [1 , 3 , 2].

Your task is to find the maximum number of elements Eshag can delete from the array a by applying the operation described above some number (maybe, zero) times.

A sequence b is a subsequence of an array c if b can be obtained from c by deletion of several (possibly, zero or all) elements.

Input

The first line contains an integer t (1≤ t≤ 100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (1≤ n≤ 100) — the length of the array a.

The second line of each test case contains n integers a_1, a_2, …, a_n (1≤ a_i ≤ 100) — the elements of the array a.

Output

For each test case print a single integer — the maximum number of elements Eshag can delete from the array a.

Example

Input


3
6
1 1 1 2 2 3
6
9 9 9 9 9 9
6
6 4 1 1 4 1


Output


3
0
3

Note

Consider the first test case.

Initially a = [1, 1, 1, 2, 2, 3].

In the first operation, Eshag can choose the subsequence containing a_1, a_5 and a_6, their average is equal to (a_1 + a_5 + a_6)/(3) = 6/3 = 2. So a_6 will be deleted.

After this a = [1, 1, 1, 2, 2].

In the second operation, Eshag can choose the subsequence containing the whole array a, the average of all its elements is equal to 7/5. So a_4 and a_5 will be deleted.

After this a = [1, 1, 1].

In the second test case, Eshag can't delete any element.
//WHEN IN DOUBT , USE BRUTE FORCE !!!!!!!!!
import java.io.*;
import java.util.*;
//class Main    //AtCoder
//class Solution // Codechef
public class Solution2 //Codeforces
{
    public static void main(String args[])
    {
        try {
            FastReader sc = new FastReader();
            int TT = sc.nextInt();
            for(int hehe=1 ; hehe  <= TT ; hehe++) {
                int N=sc.nextInt();
                int arr[]=new int[N];
                HashMap<Integer,Integer> hm=new HashMap<>();
                for(int i=0;i<N;i++) {
                    arr[i] = sc.nextInt();
                    hm.put(arr[i],hm.getOrDefault(arr[i],0)+1);
                }
                Arrays.sort(arr);
                sopln(N-hm.getOrDefault(arr[0],0));
                out.flush();
            }
        }catch(Exception e){
            sopln(e.toString());
        }
    }

    static void ruffleSort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static void ruffleSortRev(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);Collections.reverse(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }

    static int power(int x, int y, int p)
    {

        // Initialize result
        int res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static int modInverse(int n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static int nCrModPFermat(int n, int r,
                             int p)
    {

        if (n<r)
            return 0;
        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        int[] fac = new int[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;

        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
                % p;
    }


    static boolean isPrime(long n)
    {

        // Check if number is less than
        // equal to 1
        if (n <= 1)
            return false;

            // Check if number is 2
        else if (n == 2)
            return true;

            // Check if n is a multiple of 2
        else if (n % 2 == 0)
            return false;

        // If not, then just check the odds
        for (int i = 3; i <= Math.sqrt(n); i += 2)
        {
            if (n % i == 0)
                return false;
        }
        return true;
    }

    static long gcd(long a, long b)
    {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }

    static long countDig(long N){
        if(N < 10)
            return 1;
        else
            return 1+countDig(N/10);
    }

    private static boolean check(long[] arr) {
        for(int i=0;i<arr.length;i++){
            if(arr[i]!=0)
                return false;
        }
        return true;
    }

    public final static int d = 256;
    static int MOD = 1000000007;
    static final double PI = Math.PI;
    private static BufferedReader in = new BufferedReader (new InputStreamReader (System.in));
    private static PrintWriter out = new PrintWriter (new OutputStreamWriter (System.out));
    static class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        char nextChar()
        {
            try {
                return (char) (br.read());
            }catch (IOException e){
                return '~';
            }
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
    static void sop(Object o){System.out.print(o);}
    static double ceil(double d){
        return Math.ceil(d);
    }
    static double floor(double d){
        return Math.floor(d);
    }
    static double round(double d){
        return Math.round(d);
    }

    static void sopln(Object o){System.out.println(o);}
    static void printArray(boolean[] arr){
        for(int i=0;i<arr.length;i++){
            sop(arr[i]+" ");
        }
        sopln("");
    }
    static void printArray(long[] arr){
        for(int i=0;i<arr.length;i++){
            sop(arr[i]+" ");
        }
        sopln("");
    }
    static long power(long x, long y, long p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, int p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.

    ///

    // Function to find modular
// inverse of a under modulo p
// using Fermat's method.
// Assumption: p is prime
    static long modFact(int n,
                        int p)
    {
        if (n >= p)
            return 0;

        long result = 1;
        for (int i = 1; i <= n; i++)
            result = (result * i) % p;

        return result;
    }

}