#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long mod_add(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a + b) % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long ceil_div(long long a, long long b) {
  return a % b == 0 ? a / b : a / b + 1;
}
long long pwr(long long a, long long b) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, j, ans, temp, sum, k;
  string sans;
  t = 1;
  cin >> t;
  while (t--) {
    sans = "NO";
    ans = temp = sum = 0;
    cin >> n >> k;
    vector<long long> a(n + 1, 0);
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    if (sum <= k) {
      cout << 0 << "\n";
      continue;
    }
    if (sum == k + 1) {
      cout << 1 << "\n";
      continue;
    }
    sort((a).begin() + 1, (a).end());
    long long diff = sum - k;
    ans = sum - k;
    sum = 0;
    for (i = n; i > 1; i--) {
      sum += a[i];
      long long x;
      if (sum - (n - i + 1) * a[1] >= diff)
        x = 0;
      else
        x = max(ceil_div(diff - sum + (n - i + 1) * a[1], n - i + 2), 0LL);
      ans = min(ans, x + n - i + 1);
    }
    cout << ans << "\n";
  }
  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;
const long long mod = 1e9 + 7;
long long inv(long long i) {
  if (i == 1) return 1;
  return (mod - ((mod / i) * inv(mod % i)) % mod) % mod;
}
long long mod_mul(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a * b) % mod) + mod) % mod;
}
long long mod_add(long long a, long long b) {
  a = a % mod;
  b = b % mod;
  return (((a + b) % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long ceil_div(long long a, long long b) {
  return a % b == 0 ? a / b : a / b + 1;
}
long long pwr(long long a, long long b) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t, n, i, j, ans, temp, sum, k;
  string sans;
  t = 1;
  cin >> t;
  while (t--) {
    sans = "NO";
    ans = temp = sum = 0;
    cin >> n >> k;
    vector<long long> a(n + 1, 0);
    for (i = 1; i <= n; i++) {
      cin >> a[i];
      sum += a[i];
    }
    if (sum <= k) {
      cout << 0 << "\n";
      continue;
    }
    if (sum == k + 1) {
      cout << 1 << "\n";
      continue;
    }
    sort((a).begin() + 1, (a).end());
    long long diff = sum - k;
    ans = sum - k;
    sum = 0;
    for (i = n; i > 1; i--) {
      sum += a[i];
      long long x;
      if (sum - (n - i + 1) * a[1] >= diff)
        x = 0;
      else
        x = max(ceil_div(diff - sum + (n - i + 1) * a[1], n - i + 2), 0LL);
      ans = min(ans, x + n - i + 1);
    }
    cout << ans << "\n";
  }
  return 0;
}

```