#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define SZ(a) (int)((a).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef double db;
typedef pair <int, int> Pii;
typedef pair <ll, int> Pli;
typedef unsigned long long ull;
const double eps = 1e-8;
const int mod = (int)(1e9) + 7;
const int inf = 0x7fffffff;
const int N = (int)(2e5) + 7;

int _, n, a[N];

// 值为 v 的有 cnt 个
struct node {
    int v, cnt;
    bool operator< (const node& ot) const {
        return cnt < ot.cnt;
    }
} nd[N];
int len;

// 数目为 cnt 的有 t 块
struct node1 {
    int cnt, t;
} nd1[N];
int len1;

// 后缀和 cnt_i * t_i, t_i
int mul[N], bck[N];

int main() {

    // freopen("in.in", "r", stdin);
    for (scanf("%d", &_); _; --_) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        sort(a + 1, a + 1 + n);
        len = 0;
        for (int i = 1; i <= n; ++i) {
            if (a[i] != a[i - 1]) nd[++len].v = a[i], nd[len].cnt = 1;
            else ++nd[len].cnt;
        }
        sort(nd + 1, nd + 1 + len);
        
        // puts("nd[i]: ");
        // for (int i = 1; i <= len; ++i) {
        //     printf("(%d, %d)\n", nd[i].v, nd[i].cnt);
        // }
        // puts("\n-----------");

        len1 = 0;
        nd[0].cnt = 0;
        for (int i = 1; i <= len; ++i) {
            if (nd[i - 1].cnt != nd[i].cnt) 
                nd1[++len1].cnt = nd[i].cnt, nd1[len1].t = 1;
            else ++nd1[len1].t;
        }

        // puts("nd1[i]: ");
        // for (int i = 1; i <= len1; ++i) {
        //     printf("(%d, %d)\n", nd1[i].cnt, nd1[i].t);
        // }
        // puts("\n-----------");

        mul[len1 + 1] = bck[len1 + 1] = 0;
        for (int i = len1; i >= 1; --i) {
            mul[i] = mul[i + 1] + nd1[i].cnt * nd1[i].t;
            bck[i] = bck[i + 1] + nd1[i].t;
        }
        int pre = 0, res = n;
        for (int i = 1; i <= len1; ++i) {
            res = min(res, pre + mul[i + 1] - nd1[i].cnt * bck[i + 1]);
            pre += nd1[i].cnt * nd1[i].t;
        }
        printf("%d\n", res);
    }    

    return 0;
}
### Prompt

Please create a solution in Cpp to the following problem:

Polycarp was gifted an array a of length n. Polycarp considers an array beautiful if there exists a number C, such that each number in the array occurs either zero or C times. Polycarp wants to remove some elements from the array a to make it beautiful.

For example, if n=6 and a = [1, 3, 2, 1, 4, 2], then the following options are possible to make the array a array beautiful: 

  * Polycarp removes elements at positions 2 and 5, array a becomes equal to [1, 2, 1, 2]; 
  * Polycarp removes elements at positions 1 and 6, array a becomes equal to [3, 2, 1, 4]; 
  * Polycarp removes elements at positions 1, 2 and 6, array a becomes equal to [2, 1, 4]; 



Help Polycarp determine the minimum number of elements to remove from the array a to make it beautiful.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case consists of one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — 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 ≤ 10^9) — array a.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output one integer — the minimum number of elements that Polycarp has to remove from the array a to make it beautiful.

Example

Input


3
6
1 3 2 1 4 2
4
100 100 4 100
8
1 2 3 3 3 2 6 6


Output


2
1
2

### Response

```cpp
#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define SZ(a) (int)((a).size())
#define all(x) (x).begin(), (x).end()
using namespace std;
typedef long long ll;
typedef double db;
typedef pair <int, int> Pii;
typedef pair <ll, int> Pli;
typedef unsigned long long ull;
const double eps = 1e-8;
const int mod = (int)(1e9) + 7;
const int inf = 0x7fffffff;
const int N = (int)(2e5) + 7;

int _, n, a[N];

// 值为 v 的有 cnt 个
struct node {
    int v, cnt;
    bool operator< (const node& ot) const {
        return cnt < ot.cnt;
    }
} nd[N];
int len;

// 数目为 cnt 的有 t 块
struct node1 {
    int cnt, t;
} nd1[N];
int len1;

// 后缀和 cnt_i * t_i, t_i
int mul[N], bck[N];

int main() {

    // freopen("in.in", "r", stdin);
    for (scanf("%d", &_); _; --_) {
        scanf("%d", &n);
        for (int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
        }
        sort(a + 1, a + 1 + n);
        len = 0;
        for (int i = 1; i <= n; ++i) {
            if (a[i] != a[i - 1]) nd[++len].v = a[i], nd[len].cnt = 1;
            else ++nd[len].cnt;
        }
        sort(nd + 1, nd + 1 + len);
        
        // puts("nd[i]: ");
        // for (int i = 1; i <= len; ++i) {
        //     printf("(%d, %d)\n", nd[i].v, nd[i].cnt);
        // }
        // puts("\n-----------");

        len1 = 0;
        nd[0].cnt = 0;
        for (int i = 1; i <= len; ++i) {
            if (nd[i - 1].cnt != nd[i].cnt) 
                nd1[++len1].cnt = nd[i].cnt, nd1[len1].t = 1;
            else ++nd1[len1].t;
        }

        // puts("nd1[i]: ");
        // for (int i = 1; i <= len1; ++i) {
        //     printf("(%d, %d)\n", nd1[i].cnt, nd1[i].t);
        // }
        // puts("\n-----------");

        mul[len1 + 1] = bck[len1 + 1] = 0;
        for (int i = len1; i >= 1; --i) {
            mul[i] = mul[i + 1] + nd1[i].cnt * nd1[i].t;
            bck[i] = bck[i + 1] + nd1[i].t;
        }
        int pre = 0, res = n;
        for (int i = 1; i <= len1; ++i) {
            res = min(res, pre + mul[i + 1] - nd1[i].cnt * bck[i + 1]);
            pre += nd1[i].cnt * nd1[i].t;
        }
        printf("%d\n", res);
    }    

    return 0;
}
```