#include <bits/stdc++.h>
using namespace std;
void _print(long long int x) { cerr << x << " "; }
void _print(float x) { cerr << x << " "; }
void _print(double x) { cerr << x << " "; }
void _print(char x) { cerr << x << " "; }
void _print(string x) { cerr << x << " "; }
void _print(bool x) { cerr << x << " "; }
template <class T, class V>
void _print(pair<T, V> x) {
  cerr << "{";
  _print(x.first);
  cerr << ",";
  _print(x.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> x) {
  cerr << "[ ";
  for (auto i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << " ]";
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  long long int sum = 0;
  for (long long int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    sum += a[i];
  }
  if (sum <= k) {
    cout << 0 << endl;
    return;
  }
  if (n == 1) {
    cout << a[0] - k << endl;
    return;
  }
  sort(a.begin(), a.end(), greater<long long int>());
  for (long long int i = 1; i <= n - 1; i++) {
    a[i] += a[i - 1];
  };
  long long int ans = 1e17;
  long long int i = -1, j = sum - k + 1;
  ;
  while (i < j) {
    ;
    ;
    long long int mid = (i + j) / 2;
    long long int s = sum;
    long long int sm = a[n - 1] - a[n - 2];
    bool ok = 0;
    ;
    long long int up = min(mid - 1, n - 2);
    for (long long int t = 0; t <= up; t++) {
      if (mid <= t) break;
      ;
      ;
      if (s - a[t] + (sm - (mid - t - 1)) * (t + 2) - sm <= k) {
        ok = 1;
        break;
      }
    };
    if (ok) {
      j = mid;
      ans = min(ans, j);
    } else {
      i = mid + 1;
    }
  }
  if (ans == 1e17) {
    ans = sum - k;
  }
  cout << ans << endl;
}
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  clock_t start, end;
  start = clock();
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
  cerr << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cerr << " sec " << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 _print(long long int x) { cerr << x << " "; }
void _print(float x) { cerr << x << " "; }
void _print(double x) { cerr << x << " "; }
void _print(char x) { cerr << x << " "; }
void _print(string x) { cerr << x << " "; }
void _print(bool x) { cerr << x << " "; }
template <class T, class V>
void _print(pair<T, V> x) {
  cerr << "{";
  _print(x.first);
  cerr << ",";
  _print(x.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> x) {
  cerr << "[ ";
  for (T i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> x) {
  cerr << "[ ";
  for (auto i : x) {
    _print(i);
    cerr << " ";
  }
  cerr << " ]";
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> a(n);
  long long int sum = 0;
  for (long long int i = 0; i <= n - 1; i++) {
    cin >> a[i];
    sum += a[i];
  }
  if (sum <= k) {
    cout << 0 << endl;
    return;
  }
  if (n == 1) {
    cout << a[0] - k << endl;
    return;
  }
  sort(a.begin(), a.end(), greater<long long int>());
  for (long long int i = 1; i <= n - 1; i++) {
    a[i] += a[i - 1];
  };
  long long int ans = 1e17;
  long long int i = -1, j = sum - k + 1;
  ;
  while (i < j) {
    ;
    ;
    long long int mid = (i + j) / 2;
    long long int s = sum;
    long long int sm = a[n - 1] - a[n - 2];
    bool ok = 0;
    ;
    long long int up = min(mid - 1, n - 2);
    for (long long int t = 0; t <= up; t++) {
      if (mid <= t) break;
      ;
      ;
      if (s - a[t] + (sm - (mid - t - 1)) * (t + 2) - sm <= k) {
        ok = 1;
        break;
      }
    };
    if (ok) {
      j = mid;
      ans = min(ans, j);
    } else {
      i = mid + 1;
    }
  }
  if (ans == 1e17) {
    ans = sum - k;
  }
  cout << ans << endl;
}
signed main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  clock_t start, end;
  start = clock();
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  end = clock();
  double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
  cerr << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cerr << " sec " << endl;
  return 0;
}

```