#include <bits/stdc++.h>
#pragma GCC optimize "trapv"
using namespace std;
bool isPo2(int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
int toint(const string &s) {
  stringstream second;
  second << s;
  int x;
  second >> x;
  return x;
}
string tostring(int number) {
  stringstream second;
  second << number;
  return second.str();
}
long long xorUptoN(long long n) {
  if (n % 4 == 0)
    return n;
  else if (n % 4 == 1)
    return 1;
  else if (n % 4 == 2)
    return n + 1;
  return 0;
}
long long modExpo(long long x, long long y, int p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mod_Inv(long long a, long long p) { return modExpo(a, p - 2, p); }
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> arr(n);
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    sort(arr.begin(), arr.end());
    vector<long long> pref = arr;
    for (int i = 1; i < n; i++) {
      pref[i] += pref[i - 1];
    }
    if (pref[n - 1] < k) {
      cout << 0 << "\n";
    } else {
      long long cnt = 0;
      long long ans = pref[n - 1] - k;
      for (int i = n - 1; i > 0; i--) {
        cnt++;
        long long reqSum = k - (pref[i - 1] - pref[0]);
        long long eachVal = reqSum / (cnt + 1);
        if ((reqSum < 0) and ((reqSum % (cnt + 1)) != 0)) eachVal--;
        long long temp = cnt;
        if (arr[0] > eachVal) temp += (arr[0] - eachVal);
        ans = min(ans, temp);
      }
      cout << ans << "\n";
    }
  }
}

### Prompt

Your task is to create 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>
#pragma GCC optimize "trapv"
using namespace std;
bool isPo2(int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
int toint(const string &s) {
  stringstream second;
  second << s;
  int x;
  second >> x;
  return x;
}
string tostring(int number) {
  stringstream second;
  second << number;
  return second.str();
}
long long xorUptoN(long long n) {
  if (n % 4 == 0)
    return n;
  else if (n % 4 == 1)
    return 1;
  else if (n % 4 == 2)
    return n + 1;
  return 0;
}
long long modExpo(long long x, long long y, int p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long mod_Inv(long long a, long long p) { return modExpo(a, p - 2, p); }
long long combination(long long n, long long r, long long m, long long *fact,
                      long long *ifact) {
  long long val1 = fact[n];
  long long val2 = ifact[n - r];
  long long val3 = ifact[r];
  return (((val1 * val2) % m) * val3) % m;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<long long> arr(n);
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    sort(arr.begin(), arr.end());
    vector<long long> pref = arr;
    for (int i = 1; i < n; i++) {
      pref[i] += pref[i - 1];
    }
    if (pref[n - 1] < k) {
      cout << 0 << "\n";
    } else {
      long long cnt = 0;
      long long ans = pref[n - 1] - k;
      for (int i = n - 1; i > 0; i--) {
        cnt++;
        long long reqSum = k - (pref[i - 1] - pref[0]);
        long long eachVal = reqSum / (cnt + 1);
        if ((reqSum < 0) and ((reqSum % (cnt + 1)) != 0)) eachVal--;
        long long temp = cnt;
        if (arr[0] > eachVal) temp += (arr[0] - eachVal);
        ans = min(ans, temp);
      }
      cout << ans << "\n";
    }
  }
}

```