#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 29);
const long long lnf = (1ll << 60);
const int mod = 1e9 + 7;
int add(int a, int b, int md = mod) {
  return (a + b >= md ? a + b - md : (a + b < 0 ? a + b + md : a + b));
}
int mul(int a, int b, int md = mod) { return 1ll * a * b % md; }
int pw(int a, int b, int md = mod) {
  return (b ? mul(pw(mul(a, a, md), b / 2, md), (b & 1 ? a : 1), md) : 1);
}
const int N = 2e5 + 50;
int n;
long long k, a[N], suf[N];
bool val(long long t) {
  for (long long i = t; i >= 0; i--) {
    long long mn = a[0] - i;
    if (t - i >= n) break;
    if (suf[0] - (suf[n - t + i] - mn * (t - i) + i) <= k) return 1;
  }
  return 0;
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  memset(suf, 0, sizeof suf);
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = suf[i + 1] + a[i];
  }
  long long l = -1, r = max(suf[0] - k, 0ll);
  while (r - l > 1) {
    long long md = (l + r) / 2;
    if (val(md)) {
      r = md;
    } else {
      l = md;
    }
  }
  cout << r << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) solve();
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
const int inf = (1 << 29);
const long long lnf = (1ll << 60);
const int mod = 1e9 + 7;
int add(int a, int b, int md = mod) {
  return (a + b >= md ? a + b - md : (a + b < 0 ? a + b + md : a + b));
}
int mul(int a, int b, int md = mod) { return 1ll * a * b % md; }
int pw(int a, int b, int md = mod) {
  return (b ? mul(pw(mul(a, a, md), b / 2, md), (b & 1 ? a : 1), md) : 1);
}
const int N = 2e5 + 50;
int n;
long long k, a[N], suf[N];
bool val(long long t) {
  for (long long i = t; i >= 0; i--) {
    long long mn = a[0] - i;
    if (t - i >= n) break;
    if (suf[0] - (suf[n - t + i] - mn * (t - i) + i) <= k) return 1;
  }
  return 0;
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  memset(suf, 0, sizeof suf);
  for (int i = n - 1; i >= 0; i--) {
    suf[i] = suf[i + 1] + a[i];
  }
  long long l = -1, r = max(suf[0] - k, 0ll);
  while (r - l > 1) {
    long long md = (l + r) / 2;
    if (val(md)) {
      r = md;
    } else {
      l = md;
    }
  }
  cout << r << '\n';
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) solve();
  return 0;
}

```