#include <bits/stdc++.h>
using namespace std;
void solve(long long a[], long long n, long long k, long long sum) {
  vector<long long> v;
  long long p = INT_MAX, cnt = 0;
  for (long long i = 1; i < n; i++) v.push_back(a[i] - a[0]);
  while (sum > k && v.size()) {
    p = min(cnt + (sum - k + cnt) / (cnt + 1), p);
    sum -= v.back(), cnt++;
    v.pop_back();
  }
  if (sum > k) cnt += (sum - k + cnt) / (cnt + 1);
  cout << min(cnt, p) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n], sum = 0;
    for (auto &it : a) {
      cin >> it;
      sum += it;
    }
    sort(a, a + n);
    if (n == 1) {
      if (a[0] <= k)
        cout << "0\n";
      else
        cout << a[0] - k << "\n";
    } else
      solve(a, n, k, sum);
  }
}

### Prompt

Your task is to create 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;
void solve(long long a[], long long n, long long k, long long sum) {
  vector<long long> v;
  long long p = INT_MAX, cnt = 0;
  for (long long i = 1; i < n; i++) v.push_back(a[i] - a[0]);
  while (sum > k && v.size()) {
    p = min(cnt + (sum - k + cnt) / (cnt + 1), p);
    sum -= v.back(), cnt++;
    v.pop_back();
  }
  if (sum > k) cnt += (sum - k + cnt) / (cnt + 1);
  cout << min(cnt, p) << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n], sum = 0;
    for (auto &it : a) {
      cin >> it;
      sum += it;
    }
    sort(a, a + n);
    if (n == 1) {
      if (a[0] <= k)
        cout << "0\n";
      else
        cout << a[0] - k << "\n";
    } else
      solve(a, n, k, sum);
  }
}

```