#include <bits/stdc++.h>
using namespace std;
const int ms = 200005;
#pragma GCC optimize(3, "Ofast", "inline")
long long a[ms], f[ms];
long long z, y, n, k, jis, qs, h, ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    ans = 0;
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      ans += a[i];
    }
    int jis = 0;
    sort(a + 1, a + n + 1);
    qs = a[1];
    for (int i = 1; i <= n; i++) {
      f[i] = a[i] + f[i - 1];
    }
    if (ans <= k) {
      cout << jis << endl;
    } else {
      h = 0;
      for (int i = n; ans - h > k;) {
        h = (f[n] - f[i]) - (n - i) * qs + (a[1] - qs);
        if (ans - h <= k) {
          break;
        }
        if (i - 1 >= 1) {
          jis++;
          z = (f[n] - f[i - 1] + a[1]) - (n - (i - 1) + 1) * qs;
          y = (f[n] - f[i] + a[1]) - (n - i + 1) * (qs - 1);
          if (z > y) {
            i--;
          } else {
            qs--;
          }
        } else {
          long long qss = k / n;
          jis += (a[1] - qss) - (a[1] - qs);
          if (n * qss <= k) {
            break;
          } else {
            jis++;
            break;
          }
        }
      }
      cout << jis << endl;
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
using namespace std;
const int ms = 200005;
#pragma GCC optimize(3, "Ofast", "inline")
long long a[ms], f[ms];
long long z, y, n, k, jis, qs, h, ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    ans = 0;
    f[0] = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      ans += a[i];
    }
    int jis = 0;
    sort(a + 1, a + n + 1);
    qs = a[1];
    for (int i = 1; i <= n; i++) {
      f[i] = a[i] + f[i - 1];
    }
    if (ans <= k) {
      cout << jis << endl;
    } else {
      h = 0;
      for (int i = n; ans - h > k;) {
        h = (f[n] - f[i]) - (n - i) * qs + (a[1] - qs);
        if (ans - h <= k) {
          break;
        }
        if (i - 1 >= 1) {
          jis++;
          z = (f[n] - f[i - 1] + a[1]) - (n - (i - 1) + 1) * qs;
          y = (f[n] - f[i] + a[1]) - (n - i + 1) * (qs - 1);
          if (z > y) {
            i--;
          } else {
            qs--;
          }
        } else {
          long long qss = k / n;
          jis += (a[1] - qss) - (a[1] - qs);
          if (n * qss <= k) {
            break;
          } else {
            jis++;
            break;
          }
        }
      }
      cout << jis << endl;
    }
  }
  return 0;
}

```