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>
  #pragma GCC optimize("Ofast,unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  using ll = long long;
  using ld = long double;
  using vb = vector<bool>;
  using vvb = vector<vb>;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using vl = vector<long>;
  using vll = vector<long long>;
  using vvl = vector<vl>;
  using vc = vector<char>;
  using vvc = vector<vc>;
  using vs = vector<string>;
  using pii = pair<int, int>;
  using pll = pair<long long, long long>;
  using mii = map<int, int>;
  using mll = map<long long, long long>;
  using umii = unordered_map<int, int>;
  using umll = unordered_map<long long, long long>;
  const long long mod = 1e9 + 7, inf = 1e18;
  const int Mod_N = 10e5 + 5;
  void solveTestCases() {
    ll n, k;
    cin >> n >> k;
    vll array(n);
    ll arrSum = 0;
    for (int i = 0; i < n; i++) {
      cin >> array[i];
      arrSum += array[i];
    }
    sort(array.begin(), array.end());
    stack<ll> x;
    for (auto i = 1; i < n; i++) {
      x.push(array[i] - array[0]);
    }
    ll count = 0;
    ll Min = INT_MAX;
    while (!x.empty() && arrSum > k) {
      Min = min(Min, count + (arrSum - k + count) / (count + 1));
      arrSum -= x.top();
      x.pop();
      count++;
    }
    if (arrSum > k) {
      count += (arrSum - k + count) / (count + 1);
    }
    cout << min(Min, count) << "\n";
  }
  int main() {
    ll t = 1;
    cin >> t;
    while (t--) {
      solveTestCases();
    }
  }
  