#include <bits/stdc++.h>
using namespace std;
const long long N = (long long)(5 * 1e5 + 10);
void precision(long long a) { cout << setprecision(a) << fixed; }
long long expo(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b = b >> 1;
  }
  return res;
}
long long GCD(long long A, long long B) { return (B) ? GCD(B, A % B) : A; }
long long LCM(long long A, long long B) { return A * B / GCD(A, B); }
long long max(long long A, long long B) { return (A >= B) ? A : B; }
long long min(long long A, long long B) { return (A <= B) ? A : B; }
long long ceil_div(long long a, long long b) {
  return a % b == 0 ? a / b : a / b + 1;
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) {
    cin >> it;
  }
  return istream;
}
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void sol() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  cin >> a;
  sort(a.begin(), a.end());
  vector<long long> pre(n + 1);
  for (long long i = 1; i < n; i++) pre[i] = pre[i - 1] + a[i];
  long long ans = INT_MAX;
  for (long long i = n - 1; i >= 0; i--) {
    long long sum = pre[i];
    long long cnt = n - i;
    long long req = k - sum;
    req = req / cnt;
    for (long long x = req - 5; x <= req + 5; x++) {
      if (sum + cnt * x > k) continue;
      long long curr = 0;
      if (x <= a[0]) curr += a[0] - x;
      curr += cnt - 1;
      ans = min(ans, curr);
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long t = 1;
  cin >> t;
  while (t--) {
    sol();
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const long long N = (long long)(5 * 1e5 + 10);
void precision(long long a) { cout << setprecision(a) << fixed; }
long long expo(long long a, long long b, long long m) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b = b >> 1;
  }
  return res;
}
long long GCD(long long A, long long B) { return (B) ? GCD(B, A % B) : A; }
long long LCM(long long A, long long B) { return A * B / GCD(A, B); }
long long max(long long A, long long B) { return (A >= B) ? A : B; }
long long min(long long A, long long B) { return (A <= B) ? A : B; }
long long ceil_div(long long a, long long b) {
  return a % b == 0 ? a / b : a / b + 1;
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) {
    cin >> it;
  }
  return istream;
}
bool isPrime(long long n) {
  if (n == 1) return false;
  if (n == 2) return true;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void sol() {
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  cin >> a;
  sort(a.begin(), a.end());
  vector<long long> pre(n + 1);
  for (long long i = 1; i < n; i++) pre[i] = pre[i - 1] + a[i];
  long long ans = INT_MAX;
  for (long long i = n - 1; i >= 0; i--) {
    long long sum = pre[i];
    long long cnt = n - i;
    long long req = k - sum;
    req = req / cnt;
    for (long long x = req - 5; x <= req + 5; x++) {
      if (sum + cnt * x > k) continue;
      long long curr = 0;
      if (x <= a[0]) curr += a[0] - x;
      curr += cnt - 1;
      ans = min(ans, curr);
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
  long long t = 1;
  cin >> t;
  while (t--) {
    sol();
  }
  return 0;
}

```