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 _print(long long int x) { cerr << x << " "; }
  void _print(float x) { cerr << x << " "; }
  void _print(double x) { cerr << x << " "; }
  void _print(char x) { cerr << x << " "; }
  void _print(string x) { cerr << x << " "; }
  void _print(bool x) { cerr << x << " "; }
  template <class T, class V>
  void _print(pair<T, V> x) {
    cerr << "{";
    _print(x.first);
    cerr << ",";
    _print(x.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> x) {
    cerr << "[ ";
    for (T i : x) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> x) {
    cerr << "[ ";
    for (T i : x) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> x) {
    cerr << "[ ";
    for (T i : x) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> x) {
    cerr << "[ ";
    for (auto i : x) {
      _print(i);
      cerr << " ";
    }
    cerr << " ]";
  }
  void solve() {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> a(n);
    long long int sum = 0;
    for (long long int i = 0; i <= n - 1; i++) {
      cin >> a[i];
      sum += a[i];
    }
    if (sum <= k) {
      cout << 0 << endl;
      return;
    }
    if (n == 1) {
      cout << a[0] - k << endl;
      return;
    }
    sort(a.begin(), a.end(), greater<long long int>());
    for (long long int i = 1; i <= n - 1; i++) {
      a[i] += a[i - 1];
    };
    long long int ans = 1e17;
    long long int i = -1, j = sum - k + 1;
    ;
    while (i < j) {
      ;
      ;
      long long int mid = (i + j) / 2;
      long long int s = sum;
      long long int sm = a[n - 1] - a[n - 2];
      bool ok = 0;
      ;
      long long int up = min(mid - 1, n - 2);
      for (long long int t = 0; t <= up; t++) {
        if (mid <= t) break;
        ;
        ;
        if (s - a[t] + (sm - (mid - t - 1)) * (t + 2) - sm <= k) {
          ok = 1;
          break;
        }
      };
      if (ok) {
        j = mid;
        ans = min(ans, j);
      } else {
        i = mid + 1;
      }
    }
    if (ans == 1e17) {
      ans = sum - k;
    }
    cout << ans << endl;
  }
  signed main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(NULL);
    clock_t start, end;
    start = clock();
    long long int t;
    cin >> t;
    while (t--) {
      solve();
    }
    end = clock();
    double time_taken = double(end - start) / double(CLOCKS_PER_SEC);
    cerr << "Time taken by program is : " << fixed << time_taken
         << setprecision(5);
    cerr << " sec " << endl;
    return 0;
  }
  