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;
  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;
  }
  