#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long maxn = 2e5 + 9;
void init() {}
vector<long long> preSum;
bool solve(vector<long long>& arr, long long x, long long k, long long sum) {
  if (sum - x <= k) {
    return true;
  }
  long long tSum = sum;
  long long n = arr.size();
  if (x >= n - 1) {
    if (arr[n - 1] * n <= k) {
      return true;
    }
  }
  for (long long i = n - 2; i >= 0; --i) {
    long long remMoves = x - (i + 1);
    if (remMoves < 0) {
      continue;
    }
    long long smallestNum = arr[n - 1] - remMoves;
    long long newSum = tSum - preSum[i] - arr[n - 1] + smallestNum * (i + 2);
    if (newSum <= k) {
      return true;
    }
  }
  return false;
}
void test_case(long long tc) {
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  long long sum = 0;
  for (long long i = 0; i <= n - 1; ++i) {
    cin >> arr[i];
    sum += arr[i];
  }
  preSum = vector<long long>(n, 0);
  sort(arr.begin(), arr.end(), greater<long long>());
  preSum[0] = arr[0];
  for (long long i = 1; i <= n - 1; ++i) preSum[i] = preSum[i - 1] + arr[i];
  long long start = 0, end = 1e18;
  long long ans = 0;
  while (end >= start) {
    long long mid = start + (end - start) / 2;
    if (solve(arr, mid, k, sum)) {
      ans = mid;
      end = mid - 1;
    } else {
      start = mid + 1;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long tc = 1; tc <= t; ++tc) test_case(tc);
}

### 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;
const long long mod = 1e9 + 7;
const long long maxn = 2e5 + 9;
void init() {}
vector<long long> preSum;
bool solve(vector<long long>& arr, long long x, long long k, long long sum) {
  if (sum - x <= k) {
    return true;
  }
  long long tSum = sum;
  long long n = arr.size();
  if (x >= n - 1) {
    if (arr[n - 1] * n <= k) {
      return true;
    }
  }
  for (long long i = n - 2; i >= 0; --i) {
    long long remMoves = x - (i + 1);
    if (remMoves < 0) {
      continue;
    }
    long long smallestNum = arr[n - 1] - remMoves;
    long long newSum = tSum - preSum[i] - arr[n - 1] + smallestNum * (i + 2);
    if (newSum <= k) {
      return true;
    }
  }
  return false;
}
void test_case(long long tc) {
  long long n, k;
  cin >> n >> k;
  vector<long long> arr(n);
  long long sum = 0;
  for (long long i = 0; i <= n - 1; ++i) {
    cin >> arr[i];
    sum += arr[i];
  }
  preSum = vector<long long>(n, 0);
  sort(arr.begin(), arr.end(), greater<long long>());
  preSum[0] = arr[0];
  for (long long i = 1; i <= n - 1; ++i) preSum[i] = preSum[i - 1] + arr[i];
  long long start = 0, end = 1e18;
  long long ans = 0;
  while (end >= start) {
    long long mid = start + (end - start) / 2;
    if (solve(arr, mid, k, sum)) {
      ans = mid;
      end = mid - 1;
    } else {
      start = mid + 1;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long t = 1;
  cin >> t;
  for (long long tc = 1; tc <= t; ++tc) test_case(tc);
}

```