#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
int T, n, A[maxn];
long long k, SUM[maxn];
int main() {
  scanf("%d", &T);
  while (T--) {
    int pos = 1;
    long long sum = 0, ans = 0;
    scanf("%d%lld", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &A[i]);
      sum += 1ll * A[i];
    }
    if (sum <= k) {
      puts("0");
      continue;
    }
    sort(A + 1, A + 1 + n), SUM[0] = 0;
    for (int i = 1; i <= n; i++) SUM[i] = SUM[i - 1] + 1ll * A[i];
    while (A[pos] == A[1]) pos++;
    for (int i = 1; i <= n; i++) {
      long long s = k - (SUM[i] - SUM[1]);
      if (s > 0)
        s = s / (n - i + 1);
      else
        s = -1ll * (abs(s) / (n - i + 1) + (abs(s) % (n - i + 1) != 0));
      if (s > A[1]) s = A[1];
      if (i == 1) {
        if (A[1] == s)
          ans = n - pos + 1;
        else
          ans = A[1] - s + (n - 1);
      } else {
        if (A[1] == s) {
          if (pos > i)
            ans = min(ans, 1ll * n - pos + 1);
          else
            ans = min(ans, 1ll * n - i);
        } else
          ans = min(ans, A[1] - s + (n - i));
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 maxn = 2e5 + 5;
int T, n, A[maxn];
long long k, SUM[maxn];
int main() {
  scanf("%d", &T);
  while (T--) {
    int pos = 1;
    long long sum = 0, ans = 0;
    scanf("%d%lld", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &A[i]);
      sum += 1ll * A[i];
    }
    if (sum <= k) {
      puts("0");
      continue;
    }
    sort(A + 1, A + 1 + n), SUM[0] = 0;
    for (int i = 1; i <= n; i++) SUM[i] = SUM[i - 1] + 1ll * A[i];
    while (A[pos] == A[1]) pos++;
    for (int i = 1; i <= n; i++) {
      long long s = k - (SUM[i] - SUM[1]);
      if (s > 0)
        s = s / (n - i + 1);
      else
        s = -1ll * (abs(s) / (n - i + 1) + (abs(s) % (n - i + 1) != 0));
      if (s > A[1]) s = A[1];
      if (i == 1) {
        if (A[1] == s)
          ans = n - pos + 1;
        else
          ans = A[1] - s + (n - 1);
      } else {
        if (A[1] == s) {
          if (pos > i)
            ans = min(ans, 1ll * n - pos + 1);
          else
            ans = min(ans, 1ll * n - i);
        } else
          ans = min(ans, A[1] - s + (n - i));
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```