#include <bits/stdc++.h>
long long min(long long a, long long b) { return a < b ? a : b; }
unsigned int X;
void srand_() {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  X = tv.tv_sec ^ tv.tv_usec | 1;
}
int rand_() { return (X *= 3) >> 1; }
void sort(int *aa, int l, int r) {
  while (l < r) {
    int i = l, j = l, k = r, a = aa[l + rand_() % (r - l)], tmp;
    while (j < k)
      if (aa[j] == a)
        j++;
      else if (aa[j] < a) {
        tmp = aa[i], aa[i] = aa[j], aa[j] = tmp;
        i++, j++;
      } else {
        k--;
        tmp = aa[j], aa[j] = aa[k], aa[k] = tmp;
      }
    sort(aa, l, i);
    l = k;
  }
}
int main() {
  int t;
  srand_();
  scanf("%d", &t);
  while (t--) {
    static int aa[200000];
    int n, i;
    long long sum, ans;
    scanf("%d %lld", &n, &sum);
    sum = -sum;
    for (i = 0; i < n; i++) {
      scanf("%d", &aa[i]);
      sum += aa[i];
    }
    if (sum <= 0) {
      printf("0\n");
      continue;
    }
    sort(aa, 0, n);
    ans = sum;
    for (i = n - 1; i >= 1; i--) {
      sum -= aa[i] - aa[0];
      ans = min(ans, n - i + (sum <= 0 ? 0 : (sum + n - i) / (n - i + 1)));
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

You are given an integer array a_1, a_2, ..., a_n and integer k.

In one step you can 

  * either choose some index i and decrease a_i by one (make a_i = a_i - 1); 
  * or choose two indices i and j and set a_i equal to a_j (make a_i = a_j). 



What is the minimum number of steps you need to make the sum of array ∑_{i=1}^{n}{a_i} ≤ k? (You are allowed to make values of array negative).

Input

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

The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^{15}) — the size of array a and upper bound on its sum.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the array itself.

It's guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print one integer — the minimum number of steps to make ∑_{i=1}^{n}{a_i} ≤ k.

Example

Input


4
1 10
20
2 69
6 9
7 8
1 2 1 3 1 2 1
10 1
1 2 3 1 2 6 1 6 8 10


Output


10
0
2
7

Note

In the first test case, you should decrease a_1 10 times to get the sum lower or equal to k = 10.

In the second test case, the sum of array a is already less or equal to 69, so you don't need to change it.

In the third test case, you can, for example: 

  1. set a_4 = a_3 = 1; 
  2. decrease a_4 by one, and get a_4 = 0. 

As a result, you'll get array [1, 2, 1, 0, 1, 2, 1] with sum less or equal to 8 in 1 + 1 = 2 steps.

In the fourth test case, you can, for example: 

  1. choose a_7 and decrease in by one 3 times; you'll get a_7 = -2; 
  2. choose 4 elements a_6, a_8, a_9 and a_{10} and them equal to a_7 = -2. 

As a result, you'll get array [1, 2, 3, 1, 2, -2, -2, -2, -2, -2] with sum less or equal to 1 in 3 + 4 = 7 steps.

### Response

```cpp
#include <bits/stdc++.h>
long long min(long long a, long long b) { return a < b ? a : b; }
unsigned int X;
void srand_() {
  struct timeval tv;
  gettimeofday(&tv, NULL);
  X = tv.tv_sec ^ tv.tv_usec | 1;
}
int rand_() { return (X *= 3) >> 1; }
void sort(int *aa, int l, int r) {
  while (l < r) {
    int i = l, j = l, k = r, a = aa[l + rand_() % (r - l)], tmp;
    while (j < k)
      if (aa[j] == a)
        j++;
      else if (aa[j] < a) {
        tmp = aa[i], aa[i] = aa[j], aa[j] = tmp;
        i++, j++;
      } else {
        k--;
        tmp = aa[j], aa[j] = aa[k], aa[k] = tmp;
      }
    sort(aa, l, i);
    l = k;
  }
}
int main() {
  int t;
  srand_();
  scanf("%d", &t);
  while (t--) {
    static int aa[200000];
    int n, i;
    long long sum, ans;
    scanf("%d %lld", &n, &sum);
    sum = -sum;
    for (i = 0; i < n; i++) {
      scanf("%d", &aa[i]);
      sum += aa[i];
    }
    if (sum <= 0) {
      printf("0\n");
      continue;
    }
    sort(aa, 0, n);
    ans = sum;
    for (i = n - 1; i >= 1; i--) {
      sum -= aa[i] - aa[0];
      ans = min(ans, n - i + (sum <= 0 ? 0 : (sum + n - i) / (n - i + 1)));
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```