description: 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.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int maxn = 2e5 + 5;
  int T, n, A[maxn];
  long long k, SUM[maxn];
  int main() {
    scanf("%d", &T);
    while (T--) {
      int pos = 1;
      long long sum = 0, ans = 0;
      scanf("%d%lld", &n, &k);
      for (int i = 1; i <= n; i++) {
        scanf("%d", &A[i]);
        sum += 1ll * A[i];
      }
      if (sum <= k) {
        puts("0");
        continue;
      }
      sort(A + 1, A + 1 + n), SUM[0] = 0;
      for (int i = 1; i <= n; i++) SUM[i] = SUM[i - 1] + 1ll * A[i];
      while (A[pos] == A[1]) pos++;
      for (int i = 1; i <= n; i++) {
        long long s = k - (SUM[i] - SUM[1]);
        if (s > 0)
          s = s / (n - i + 1);
        else
          s = -1ll * (abs(s) / (n - i + 1) + (abs(s) % (n - i + 1) != 0));
        if (s > A[1]) s = A[1];
        if (i == 1) {
          if (A[1] == s)
            ans = n - pos + 1;
          else
            ans = A[1] - s + (n - 1);
        } else {
          if (A[1] == s) {
            if (pos > i)
              ans = min(ans, 1ll * n - pos + 1);
            else
              ans = min(ans, 1ll * n - i);
          } else
            ans = min(ans, A[1] - s + (n - i));
        }
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  