#include <bits/stdc++.h>
using namespace std;
inline long long rd() {
  long long x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int T, n;
long long ans, All, k, a[200005], sum[200005];
bool cmp(long long x, long long y) { return x < y; }
int main() {
  T = rd();
  while (T--) {
    n = rd(), k = rd(), sum[n + 1] = sum[0] = All = 0, ans = 0x7f7f7f7f7f7f7f7f;
    for (int i = 1; i <= n; i++) a[i] = rd(), All += a[i];
    sort(a + 1, a + n + 1, cmp);
    for (int i = 1; i <= n; i++) sum[i] = a[i] - a[1];
    for (int i = n; i >= 0; i--) sum[i] += sum[i + 1];
    if (All <= k) {
      printf("0\n");
      continue;
    }
    ans = All - k;
    for (int l = 1; l < n; l++) {
      long long tmp = (All - k - sum[l + 1] + 1ll * n - 1ll * l) /
                      (1ll * n - 1ll * l + 1ll);
      if (All - k - sum[l + 1] + 1ll * n - 1ll * l <= 0) tmp = 0;
      ans = minll(ans, tmp + 1ll * n - 1ll * l);
    }
    printf("%lld\n", ans);
  }
  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;
inline long long rd() {
  long long x = 0;
  char ch, t = 0;
  while (!isdigit(ch = getchar())) t |= ch == '-';
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return x = t ? -x : x;
}
inline long long maxll(long long x, long long y) { return x > y ? x : y; }
inline long long minll(long long x, long long y) { return x < y ? x : y; }
inline long long absll(long long x) { return x > 0ll ? x : -x; }
inline long long gcd(long long x, long long y) {
  return (y == 0) ? x : gcd(y, x % y);
}
int T, n;
long long ans, All, k, a[200005], sum[200005];
bool cmp(long long x, long long y) { return x < y; }
int main() {
  T = rd();
  while (T--) {
    n = rd(), k = rd(), sum[n + 1] = sum[0] = All = 0, ans = 0x7f7f7f7f7f7f7f7f;
    for (int i = 1; i <= n; i++) a[i] = rd(), All += a[i];
    sort(a + 1, a + n + 1, cmp);
    for (int i = 1; i <= n; i++) sum[i] = a[i] - a[1];
    for (int i = n; i >= 0; i--) sum[i] += sum[i + 1];
    if (All <= k) {
      printf("0\n");
      continue;
    }
    ans = All - k;
    for (int l = 1; l < n; l++) {
      long long tmp = (All - k - sum[l + 1] + 1ll * n - 1ll * l) /
                      (1ll * n - 1ll * l + 1ll);
      if (All - k - sum[l + 1] + 1ll * n - 1ll * l <= 0) tmp = 0;
      ans = minll(ans, tmp + 1ll * n - 1ll * l);
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```