#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using ll = long long;
using ld = long double;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<long>;
using vll = vector<long long>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using mii = map<int, int>;
using mll = map<long long, long long>;
using umii = unordered_map<int, int>;
using umll = unordered_map<long long, long long>;
const long long mod = 1e9 + 7, inf = 1e18;
const int Mod_N = 10e5 + 5;
void solveTestCases() {
  ll n, k;
  cin >> n >> k;
  vll array(n);
  ll arrSum = 0;
  for (int i = 0; i < n; i++) {
    cin >> array[i];
    arrSum += array[i];
  }
  sort(array.begin(), array.end());
  stack<ll> x;
  for (auto i = 1; i < n; i++) {
    x.push(array[i] - array[0]);
  }
  ll count = 0;
  ll Min = INT_MAX;
  while (!x.empty() && arrSum > k) {
    Min = min(Min, count + (arrSum - k + count) / (count + 1));
    arrSum -= x.top();
    x.pop();
    count++;
  }
  if (arrSum > k) {
    count += (arrSum - k + count) / (count + 1);
  }
  cout << min(Min, count) << "\n";
}
int main() {
  ll t = 1;
  cin >> t;
  while (t--) {
    solveTestCases();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using ll = long long;
using ld = long double;
using vb = vector<bool>;
using vvb = vector<vb>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<long>;
using vll = vector<long long>;
using vvl = vector<vl>;
using vc = vector<char>;
using vvc = vector<vc>;
using vs = vector<string>;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using mii = map<int, int>;
using mll = map<long long, long long>;
using umii = unordered_map<int, int>;
using umll = unordered_map<long long, long long>;
const long long mod = 1e9 + 7, inf = 1e18;
const int Mod_N = 10e5 + 5;
void solveTestCases() {
  ll n, k;
  cin >> n >> k;
  vll array(n);
  ll arrSum = 0;
  for (int i = 0; i < n; i++) {
    cin >> array[i];
    arrSum += array[i];
  }
  sort(array.begin(), array.end());
  stack<ll> x;
  for (auto i = 1; i < n; i++) {
    x.push(array[i] - array[0]);
  }
  ll count = 0;
  ll Min = INT_MAX;
  while (!x.empty() && arrSum > k) {
    Min = min(Min, count + (arrSum - k + count) / (count + 1));
    arrSum -= x.top();
    x.pop();
    count++;
  }
  if (arrSum > k) {
    count += (arrSum - k + count) / (count + 1);
  }
  cout << min(Min, count) << "\n";
}
int main() {
  ll t = 1;
  cin >> t;
  while (t--) {
    solveTestCases();
  }
}

```