#include <bits/stdc++.h>
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(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...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        std::chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const long long INF = 9223372036854775807;
const long long N = 2e5 + 5;
const long long mod = 998244353;
const long long M = 22;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 2];
    for (long long i = 0; i <= n - 1; i++) cin >> a[i];
    sort(a, a + n);
    vector<long long> v;
    long long sum = 0;
    for (long long i = n - 1; i >= 1; i--) {
      sum += a[i];
      v.push_back(sum);
    }
    sum += a[0];
    long long ans = 0;
    if (sum <= k)
      cout << "0" << '\n';
    else {
      if (a[0] > (k / n)) {
        ans += (a[0] - (k / n));
        sum -= ans;
        a[0] = k / n;
      }
      if (sum <= k) {
        cout << ans << '\n';
        continue;
      }
      long long p = INF;
      for (long long i = 0; i <= n; i++) {
        long long l = 0, r = n - 2, z = n - 1;
        while (r >= l) {
          long long mid = (l + r) / 2;
          long long ns = (sum - i) - (v[mid] - ((mid + 1) * (a[0] - i)));
          if (ns <= k) {
            z = mid;
            r = mid - 1;
          } else
            l = mid + 1;
        }
        p = min(p, z + i + 1);
      }
      cout << ans + p << '\n';
    }
  }
}

### 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>
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(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...);
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        std::chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
const long long INF = 9223372036854775807;
const long long N = 2e5 + 5;
const long long mod = 998244353;
const long long M = 22;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long a[n + 2];
    for (long long i = 0; i <= n - 1; i++) cin >> a[i];
    sort(a, a + n);
    vector<long long> v;
    long long sum = 0;
    for (long long i = n - 1; i >= 1; i--) {
      sum += a[i];
      v.push_back(sum);
    }
    sum += a[0];
    long long ans = 0;
    if (sum <= k)
      cout << "0" << '\n';
    else {
      if (a[0] > (k / n)) {
        ans += (a[0] - (k / n));
        sum -= ans;
        a[0] = k / n;
      }
      if (sum <= k) {
        cout << ans << '\n';
        continue;
      }
      long long p = INF;
      for (long long i = 0; i <= n; i++) {
        long long l = 0, r = n - 2, z = n - 1;
        while (r >= l) {
          long long mid = (l + r) / 2;
          long long ns = (sum - i) - (v[mid] - ((mid + 1) * (a[0] - i)));
          if (ns <= k) {
            z = mid;
            r = mid - 1;
          } else
            l = mid + 1;
        }
        p = min(p, z + i + 1);
      }
      cout << ans + p << '\n';
    }
  }
}

```