#include <bits/stdc++.h>
using namespace std;
inline void smax(int& a, int b) {
  if (b > a) a = b;
}
inline void smin(int& a, int b) {
  if (b < a) a = b;
}
inline void smax(long long& a, long long b) {
  if (b > a) a = b;
}
inline void smin(long long& a, long long b) {
  if (b < a) a = b;
}
const long long LLINF = 2e18 + 13;
const int INF = 1e9 + 7;
mt19937 rng(time(nullptr) + 'm' + 'a' + 'n' + 'u' + 'l');
uniform_int_distribution<int> uid(0, INF);
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n);
    vector<long long> pr(n + 1, 0);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < n; ++i) pr[i + 1] = pr[i] + a[i];
    long long sum = pr[n];
    if (sum <= k) {
      cout << 0 << endl;
      continue;
    }
    long long ans = sum - k;
    for (int i = 1; i < n; ++i) {
      long long minus = pr[i];
      long long lans = n - i;
      long long l = -1, r = INF * 10ll;
      while (r - l > 1) {
        long long m = r + l >> 1;
        if (minus - m + (a[0] - m) * lans <= k) {
          r = m;
        } else {
          l = m;
        }
      }
      smin(ans, lans + r);
    }
    cout << ans << endl;
  NEXT:;
  }
  return 0;
}

### Prompt

Please formulate 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;
inline void smax(int& a, int b) {
  if (b > a) a = b;
}
inline void smin(int& a, int b) {
  if (b < a) a = b;
}
inline void smax(long long& a, long long b) {
  if (b > a) a = b;
}
inline void smin(long long& a, long long b) {
  if (b < a) a = b;
}
const long long LLINF = 2e18 + 13;
const int INF = 1e9 + 7;
mt19937 rng(time(nullptr) + 'm' + 'a' + 'n' + 'u' + 'l');
uniform_int_distribution<int> uid(0, INF);
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n);
    vector<long long> pr(n + 1, 0);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    for (int i = 0; i < n; ++i) pr[i + 1] = pr[i] + a[i];
    long long sum = pr[n];
    if (sum <= k) {
      cout << 0 << endl;
      continue;
    }
    long long ans = sum - k;
    for (int i = 1; i < n; ++i) {
      long long minus = pr[i];
      long long lans = n - i;
      long long l = -1, r = INF * 10ll;
      while (r - l > 1) {
        long long m = r + l >> 1;
        if (minus - m + (a[0] - m) * lans <= k) {
          r = m;
        } else {
          l = m;
        }
      }
      smin(ans, lans + r);
    }
    cout << ans << endl;
  NEXT:;
  }
  return 0;
}

```