#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long k;
    cin >> k;
    long long a[n], i;
    long long sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    sort(a, a + n);
    long long c = 0;
    for (i = 0; i < n; i++) {
      if (a[i] == a[0]) c++;
    }
    long long ans = 1e12;
    if (n == 1) {
      if (a[0] > k)
        ans = a[0] - k;
      else
        ans = 0;
      cout << ans << "\n";
      continue;
    }
    if (sum <= k) {
      cout << 0 << "\n";
      continue;
    }
    long long pre[n];
    pre[0] = a[0];
    for (i = 1; i < n; i++) {
      pre[i] = pre[i - 1] + a[i];
    }
    for (i = 0; i <= n - 1; i++) {
      sum = pre[n - i - 1] - pre[0];
      long long d = k - sum;
      d = floor((double)d / (i + 1));
      if (d > a[0]) {
        ans = min(ans, i);
        continue;
      }
      long long move = a[0] - d;
      move += i;
      ans = min(ans, move);
    }
    cout << ans << "\n";
  }
  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;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long k;
    cin >> k;
    long long a[n], i;
    long long sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    sort(a, a + n);
    long long c = 0;
    for (i = 0; i < n; i++) {
      if (a[i] == a[0]) c++;
    }
    long long ans = 1e12;
    if (n == 1) {
      if (a[0] > k)
        ans = a[0] - k;
      else
        ans = 0;
      cout << ans << "\n";
      continue;
    }
    if (sum <= k) {
      cout << 0 << "\n";
      continue;
    }
    long long pre[n];
    pre[0] = a[0];
    for (i = 1; i < n; i++) {
      pre[i] = pre[i - 1] + a[i];
    }
    for (i = 0; i <= n - 1; i++) {
      sum = pre[n - i - 1] - pre[0];
      long long d = k - sum;
      d = floor((double)d / (i + 1));
      if (d > a[0]) {
        ans = min(ans, i);
        continue;
      }
      long long move = a[0] - d;
      move += i;
      ans = min(ans, move);
    }
    cout << ans << "\n";
  }
  return 0;
}

```