#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    long long n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n);
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    sort(a.rbegin(), a.rend());
    long long baki = sum - k;
    if (baki <= 0)
      cout << 0 << endl;
    else {
      long long baad = 0, res = baki;
      for (int i = 0; i < n - 1; i++) {
        baad += (a[i] - a[n - 1]);
        if (baad >= baki) {
          res = min(res, (long long)(i + 1));
          break;
        }
        long long x = ((baki - baad) / (i + 2));
        if ((baki - baad) % (i + 2)) x++;
        res = min(res, x + i + 1);
      }
      cout << res << endl;
    }
  }
  return 0;
}

### Prompt

Generate a Cpp solution to 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>
using namespace std;
const int N = 100005;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    long long n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n);
    long long sum = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    sort(a.rbegin(), a.rend());
    long long baki = sum - k;
    if (baki <= 0)
      cout << 0 << endl;
    else {
      long long baad = 0, res = baki;
      for (int i = 0; i < n - 1; i++) {
        baad += (a[i] - a[n - 1]);
        if (baad >= baki) {
          res = min(res, (long long)(i + 1));
          break;
        }
        long long x = ((baki - baad) / (i + 2));
        if ((baki - baad) % (i + 2)) x++;
        res = min(res, x + i + 1);
      }
      cout << res << endl;
    }
  }
  return 0;
}

```