#include <bits/stdc++.h>
using namespace std;
long long read() {
  long long ans = 0;
  char last = ' ', ch = getchar();
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
const int N = 2e5 + 5;
int n;
long long k, sum, tmp, suf[N], a[N], c, ans;
long long accCeil(long long a, long long b) {
  long long res = a / b;
  if (res * b < a) res++;
  return res;
}
int main() {
  int T = read();
  while (T--) {
    n = read(), k = read();
    sum = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      sum += a[i];
    }
    if (sum <= k) {
      printf("0\n");
      continue;
    }
    if (n == 1) {
      printf("%lld\n", sum - k);
      continue;
    }
    sort(a + 1, a + n + 1);
    suf[n + 1] = 0;
    ans = 1e18;
    for (int i = n; i > 1; --i) {
      suf[i] = suf[i + 1] + a[i];
      tmp = sum - k - suf[i] + (n - i + 1) * a[1];
      c = max(accCeil(tmp, n - i + 2), 0ll);
      ans = min(ans, n - i + 1 + c);
    }
    printf("%lld\n", min(ans, sum - k));
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long read() {
  long long ans = 0;
  char last = ' ', ch = getchar();
  while (ch < '0' || ch > '9') last = ch, ch = getchar();
  while (ch >= '0' && ch <= '9') ans = ans * 10 + ch - '0', ch = getchar();
  if (last == '-') ans = -ans;
  return ans;
}
const int N = 2e5 + 5;
int n;
long long k, sum, tmp, suf[N], a[N], c, ans;
long long accCeil(long long a, long long b) {
  long long res = a / b;
  if (res * b < a) res++;
  return res;
}
int main() {
  int T = read();
  while (T--) {
    n = read(), k = read();
    sum = 0;
    for (int i = 1; i <= n; ++i) {
      a[i] = read();
      sum += a[i];
    }
    if (sum <= k) {
      printf("0\n");
      continue;
    }
    if (n == 1) {
      printf("%lld\n", sum - k);
      continue;
    }
    sort(a + 1, a + n + 1);
    suf[n + 1] = 0;
    ans = 1e18;
    for (int i = n; i > 1; --i) {
      suf[i] = suf[i + 1] + a[i];
      tmp = sum - k - suf[i] + (n - i + 1) * a[1];
      c = max(accCeil(tmp, n - i + 2), 0ll);
      ans = min(ans, n - i + 1 + c);
    }
    printf("%lld\n", min(ans, sum - k));
  }
  return 0;
}

```