#include <bits/stdc++.h>
using namespace std;
const int64_t N = 1e5 + 5, mod = 1e9 + 7;
const int64_t N1 = 1e18;
void solve() {
  int64_t n, m;
  cin >> n >> m;
  int64_t ans = 1e15;
  vector<int64_t> z(n);
  for (int64_t &it : z) {
    cin >> it;
  }
  sort(z.begin(), z.end());
  vector<int64_t> prf(n);
  for (int64_t i = 0; i < n; i++) {
    prf[i] = z[i];
    if (i) {
      prf[i] += prf[i - 1];
    }
  }
  int64_t gh = prf.back();
  auto plz = [&](int64_t x) {
    int64_t lo = 0, hi = n - 1, mid, curans = 2e15;
    while (lo <= hi) {
      mid = (lo + hi) >> 1;
      if (gh - (prf[n - 1] - prf[n - 1 - mid]) - z[0] + x + x * mid <= m) {
        curans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    return curans;
  };
  int64_t lo = 0, hi = 1e15;
  while (hi - lo > 3) {
    int64_t mid1 = lo + (hi - lo) / 3;
    int64_t mid2 = hi - (hi - lo) / 3;
    if (plz(z[0] - mid1) >= 2e15) {
      lo = mid1;
      continue;
    }
    if (-(mid1 + plz(z[0] - mid1)) < -(mid2 + plz(z[0] - mid2))) {
      lo = mid1;
    } else {
      hi = mid2;
    }
  }
  for (int64_t cur = lo; cur <= hi + (n + 2) / 3; cur++) {
    ans = min(ans, cur + plz(z[0] - cur));
  }
  cout << ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  int64_t tq = 1;
  cin >> tq;
  for (; tq; tq--) {
    solve();
  }
}

### 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 int64_t N = 1e5 + 5, mod = 1e9 + 7;
const int64_t N1 = 1e18;
void solve() {
  int64_t n, m;
  cin >> n >> m;
  int64_t ans = 1e15;
  vector<int64_t> z(n);
  for (int64_t &it : z) {
    cin >> it;
  }
  sort(z.begin(), z.end());
  vector<int64_t> prf(n);
  for (int64_t i = 0; i < n; i++) {
    prf[i] = z[i];
    if (i) {
      prf[i] += prf[i - 1];
    }
  }
  int64_t gh = prf.back();
  auto plz = [&](int64_t x) {
    int64_t lo = 0, hi = n - 1, mid, curans = 2e15;
    while (lo <= hi) {
      mid = (lo + hi) >> 1;
      if (gh - (prf[n - 1] - prf[n - 1 - mid]) - z[0] + x + x * mid <= m) {
        curans = mid;
        hi = mid - 1;
      } else {
        lo = mid + 1;
      }
    }
    return curans;
  };
  int64_t lo = 0, hi = 1e15;
  while (hi - lo > 3) {
    int64_t mid1 = lo + (hi - lo) / 3;
    int64_t mid2 = hi - (hi - lo) / 3;
    if (plz(z[0] - mid1) >= 2e15) {
      lo = mid1;
      continue;
    }
    if (-(mid1 + plz(z[0] - mid1)) < -(mid2 + plz(z[0] - mid2))) {
      lo = mid1;
    } else {
      hi = mid2;
    }
  }
  for (int64_t cur = lo; cur <= hi + (n + 2) / 3; cur++) {
    ans = min(ans, cur + plz(z[0] - cur));
  }
  cout << ans << "\n";
}
int32_t main() {
  ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  int64_t tq = 1;
  cin >> tq;
  for (; tq; tq--) {
    solve();
  }
}

```