#include <bits/stdc++.h>
using namespace std;
vector<long long> sieve(long long n) {
  long long* arr = new long long[n + 1]();
  vector<long long> vect;
  for (long long i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long expo(long long a, long long b, long long MOD) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long findGCD(long long arr[], long long n) {
  long long result = arr[0];
  for (long long i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
string decToBinary(long long n) {
  string output = "";
  while (n) {
    if (n & 1) {
      output = '1' + output;
    } else {
      output = '0' + output;
    }
    n >>= 1;
  }
  return output;
}
long long isSubstring(string s1, string s2) {
  long long M = s1.length();
  long long N = s2.length();
  for (long long i = 0; i <= N - M; i++) {
    long long j;
    for (j = 0; j < M; j++)
      if (s2[i + j] != s1[j]) break;
    if (j == M) return i;
  }
  return -1;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long sum[n + 1];
  sum[1] = arr[0];
  for (long long i = 1; i < n; i++) {
    sum[i + 1] = sum[i] + arr[i];
  }
  long long ans = 1e18;
  for (long long i = 0; i < n; i++) {
    long long x;
    long long m = k + arr[0] - sum[n - i];
    if ((m / (i + 1)) * (i + 1) > m) {
      x = arr[0] - (m / (i + 1) - 1);
    } else {
      x = arr[0] - (m / (i + 1));
    }
    ans = min(ans, i + max(0ll, x));
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
vector<long long> sieve(long long n) {
  long long* arr = new long long[n + 1]();
  vector<long long> vect;
  for (long long i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (long long j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
long long expo(long long a, long long b, long long MOD) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % MOD;
    a = (a * a) % MOD;
    b = b >> 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long findGCD(long long arr[], long long n) {
  long long result = arr[0];
  for (long long i = 1; i < n; i++) {
    result = gcd(arr[i], result);
    if (result == 1) {
      return 1;
    }
  }
  return result;
}
string decToBinary(long long n) {
  string output = "";
  while (n) {
    if (n & 1) {
      output = '1' + output;
    } else {
      output = '0' + output;
    }
    n >>= 1;
  }
  return output;
}
long long isSubstring(string s1, string s2) {
  long long M = s1.length();
  long long N = s2.length();
  for (long long i = 0; i <= N - M; i++) {
    long long j;
    for (j = 0; j < M; j++)
      if (s2[i + j] != s1[j]) break;
    if (j == M) return i;
  }
  return -1;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  long long sum[n + 1];
  sum[1] = arr[0];
  for (long long i = 1; i < n; i++) {
    sum[i + 1] = sum[i] + arr[i];
  }
  long long ans = 1e18;
  for (long long i = 0; i < n; i++) {
    long long x;
    long long m = k + arr[0] - sum[n - i];
    if ((m / (i + 1)) * (i + 1) > m) {
      x = arr[0] - (m / (i + 1) - 1);
    } else {
      x = arr[0] - (m / (i + 1));
    }
    ans = min(ans, i + max(0ll, x));
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```