#include <bits/stdc++.h>
using namespace std;
const long long NR = 2e5 + 10;
void Min(long long& x, long long y) { x = min(x, y); }
void Max(long long& x, long long y) { x = max(x, y); }
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n, k;
long long a[NR];
void work() {
  n = read(), k = read();
  a[n + 1] = 0;
  for (long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  long long sum = 0, ans = 0x3f3f3f3f3f3f3f3fll;
  for (long long i = 2; i <= n; i++) sum += a[i];
  for (long long i = n + 1; i >= 2; i--) {
    sum -= a[i];
    long long p = k - sum;
    long long q = n - i + 2;
    if (p >= 0)
      p = p / q;
    else
      p = (p - q + 1) / q;
    ans = min(ans, n - i + 1 + max(0ll, a[1] - p));
  }
  printf("%lld\n", ans);
}
signed main() {
  long long T = read();
  while (T--) work();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const long long NR = 2e5 + 10;
void Min(long long& x, long long y) { x = min(x, y); }
void Max(long long& x, long long y) { x = max(x, y); }
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
long long n, k;
long long a[NR];
void work() {
  n = read(), k = read();
  a[n + 1] = 0;
  for (long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  long long sum = 0, ans = 0x3f3f3f3f3f3f3f3fll;
  for (long long i = 2; i <= n; i++) sum += a[i];
  for (long long i = n + 1; i >= 2; i--) {
    sum -= a[i];
    long long p = k - sum;
    long long q = n - i + 2;
    if (p >= 0)
      p = p / q;
    else
      p = (p - q + 1) / q;
    ans = min(ans, n - i + 1 + max(0ll, a[1] - p));
  }
  printf("%lld\n", ans);
}
signed main() {
  long long T = read();
  while (T--) work();
  return 0;
}

```