#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long int MOD = 1e9 + 7;
const long long int INF = 1e18;
const long long int maxn = 1e6 + 4;
long long int binpow(long long int b, long long int n, long long int mod) {
  long long int res = 1;
  while (n) {
    if (n & 1) res = res * b % mod;
    n /= 2;
    b = b * b % mod;
  }
  return res;
}
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long> pref(n + 1);
  for (int i = 0; i < n; i++) {
    pref[i + 1] = pref[i] + a[i];
  }
  long long ans = (long long)9e18;
  for (int t = 0; t <= n - 1; t++) {
    long long sum = pref[n - t] + a[0] * t;
    long long cur = t;
    if (sum > k) {
      long long diff = sum - k;
      cur += (diff + t) / (t + 1);
    }
    ans = min(ans, cur);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

### 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const long long int MOD = 1e9 + 7;
const long long int INF = 1e18;
const long long int maxn = 1e6 + 4;
long long int binpow(long long int b, long long int n, long long int mod) {
  long long int res = 1;
  while (n) {
    if (n & 1) res = res * b % mod;
    n /= 2;
    b = b * b % mod;
  }
  return res;
}
void solve() {
  int n;
  long long k;
  cin >> n >> k;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long> pref(n + 1);
  for (int i = 0; i < n; i++) {
    pref[i + 1] = pref[i] + a[i];
  }
  long long ans = (long long)9e18;
  for (int t = 0; t <= n - 1; t++) {
    long long sum = pref[n - t] + a[0] * t;
    long long cur = t;
    if (sum > k) {
      long long diff = sum - k;
      cur += (diff + t) / (t + 1);
    }
    ans = min(ans, cur);
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) solve();
}

```