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 "trapv"
  using namespace std;
  bool isPo2(int n) {
    if (n == 0) return false;
    return (ceil(log2(n)) == floor(log2(n)));
  }
  int toint(const string &s) {
    stringstream second;
    second << s;
    int x;
    second >> x;
    return x;
  }
  string tostring(int number) {
    stringstream second;
    second << number;
    return second.str();
  }
  long long xorUptoN(long long n) {
    if (n % 4 == 0)
      return n;
    else if (n % 4 == 1)
      return 1;
    else if (n % 4 == 2)
      return n + 1;
    return 0;
  }
  long long modExpo(long long x, long long y, int p) {
    long long res = 1;
    x = x % p;
    if (x == 0) return 0;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  long long mod_Inv(long long a, long long p) { return modExpo(a, p - 2, p); }
  long long combination(long long n, long long r, long long m, long long *fact,
                        long long *ifact) {
    long long val1 = fact[n];
    long long val2 = ifact[n - r];
    long long val3 = ifact[r];
    return (((val1 * val2) % m) * val3) % m;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<long long> arr(n);
      for (int i = 0; i < n; i++) {
        cin >> arr[i];
      }
      sort(arr.begin(), arr.end());
      vector<long long> pref = arr;
      for (int i = 1; i < n; i++) {
        pref[i] += pref[i - 1];
      }
      if (pref[n - 1] < k) {
        cout << 0 << "\n";
      } else {
        long long cnt = 0;
        long long ans = pref[n - 1] - k;
        for (int i = n - 1; i > 0; i--) {
          cnt++;
          long long reqSum = k - (pref[i - 1] - pref[0]);
          long long eachVal = reqSum / (cnt + 1);
          if ((reqSum < 0) and ((reqSum % (cnt + 1)) != 0)) eachVal--;
          long long temp = cnt;
          if (arr[0] > eachVal) temp += (arr[0] - eachVal);
          ans = min(ans, temp);
        }
        cout << ans << "\n";
      }
    }
  }
  