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 int64_t N = 1e5 + 5, mod = 1e9 + 7;
  const int64_t N1 = 1e18;
  void solve() {
    int64_t n, m;
    cin >> n >> m;
    int64_t ans = 1e15;
    vector<int64_t> z(n);
    for (int64_t &it : z) {
      cin >> it;
    }
    sort(z.begin(), z.end());
    vector<int64_t> prf(n);
    for (int64_t i = 0; i < n; i++) {
      prf[i] = z[i];
      if (i) {
        prf[i] += prf[i - 1];
      }
    }
    int64_t gh = prf.back();
    auto plz = [&](int64_t x) {
      int64_t lo = 0, hi = n - 1, mid, curans = 2e15;
      while (lo <= hi) {
        mid = (lo + hi) >> 1;
        if (gh - (prf[n - 1] - prf[n - 1 - mid]) - z[0] + x + x * mid <= m) {
          curans = mid;
          hi = mid - 1;
        } else {
          lo = mid + 1;
        }
      }
      return curans;
    };
    int64_t lo = 0, hi = 1e15;
    while (hi - lo > 3) {
      int64_t mid1 = lo + (hi - lo) / 3;
      int64_t mid2 = hi - (hi - lo) / 3;
      if (plz(z[0] - mid1) >= 2e15) {
        lo = mid1;
        continue;
      }
      if (-(mid1 + plz(z[0] - mid1)) < -(mid2 + plz(z[0] - mid2))) {
        lo = mid1;
      } else {
        hi = mid2;
      }
    }
    for (int64_t cur = lo; cur <= hi + (n + 2) / 3; cur++) {
      ans = min(ans, cur + plz(z[0] - cur));
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);
    int64_t tq = 1;
    cin >> tq;
    for (; tq; tq--) {
      solve();
    }
  }
  