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;
  void input(vector<int> &a, int n) {
    for (int i = 0; i < n; i += 1) cin >> a[i];
  }
  void display(vector<int> &a) {
    for (int i = 0; i < a.size(); i += 1) cout << a[i] << " ";
    cout << "\n";
  }
  bool isFeasible(vector<long long> &s, long long k, long long swaps) {
    long long low = s[0];
    long long len = s.size() - 1;
    int n = min(len, swaps);
    if (swaps > n) low -= (swaps - len);
    while (n) {
      if ((long long)(s[len - n] - s[0] + low + (long long)low * (long long)n) <=
          k)
        return true;
      low--;
      n--;
    }
    return false;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
      int n;
      long long k;
      cin >> n >> k;
      vector<int> a(n);
      input(a, n);
      long long sum = accumulate(a.begin(), a.end(), 0LL);
      if (sum <= k) {
        cout << "0\n";
        continue;
      }
      sort(a.begin(), a.end());
      long long low = 0, high = n - 1 + max(a[0] - (k / n), 0LL);
      vector<long long> s(n, 0);
      s[0] = a[0];
      for (int i = 1; i < n; i += 1) {
        s[i] = s[i - 1] + a[i];
      }
      long long mid, res = high;
      while (low <= high) {
        mid = low + (high - low) / 2;
        if (isFeasible(s, k, mid)) {
          res = mid;
          high = mid - 1;
        } else
          low = mid + 1;
      }
      cout << min(res, sum - k) << "\n";
    }
    return 0;
  }
  