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>
  long long min(long long a, long long b) { return a < b ? a : b; }
  unsigned int X;
  void srand_() {
    struct timeval tv;
    gettimeofday(&tv, NULL);
    X = tv.tv_sec ^ tv.tv_usec | 1;
  }
  int rand_() { return (X *= 3) >> 1; }
  void sort(int *aa, int l, int r) {
    while (l < r) {
      int i = l, j = l, k = r, a = aa[l + rand_() % (r - l)], tmp;
      while (j < k)
        if (aa[j] == a)
          j++;
        else if (aa[j] < a) {
          tmp = aa[i], aa[i] = aa[j], aa[j] = tmp;
          i++, j++;
        } else {
          k--;
          tmp = aa[j], aa[j] = aa[k], aa[k] = tmp;
        }
      sort(aa, l, i);
      l = k;
    }
  }
  int main() {
    int t;
    srand_();
    scanf("%d", &t);
    while (t--) {
      static int aa[200000];
      int n, i;
      long long sum, ans;
      scanf("%d %lld", &n, &sum);
      sum = -sum;
      for (i = 0; i < n; i++) {
        scanf("%d", &aa[i]);
        sum += aa[i];
      }
      if (sum <= 0) {
        printf("0\n");
        continue;
      }
      sort(aa, 0, n);
      ans = sum;
      for (i = n - 1; i >= 1; i--) {
        sum -= aa[i] - aa[0];
        ans = min(ans, n - i + (sum <= 0 ? 0 : (sum + n - i) / (n - i + 1)));
      }
      printf("%lld\n", ans);
    }
    return 0;
  }
  