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 c_p_c() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  long long calc(int a, int b) {
    if (a % b == 0)
      return a / b;
    else
      return a / b + 1;
  }
  unsigned long long power(unsigned long long x, int y, int p) {
    unsigned long long res = 1;
    x = x % p;
    while (y > 0) {
      if (y & 1) res = (res * x) % p;
      y = y >> 1;
      x = (x * x) % p;
    }
    return res;
  }
  unsigned long long modInverse(unsigned long long n, int p) {
    return power(n, p - 2, p);
  }
  long long fac[100001];
  void solve() {
    fac[0] = 1;
    for (int i = 1; i <= 100000; i++) fac[i] = (fac[i - 1] * i) % 1000000007;
  }
  unsigned long long nCrModPFermat(unsigned long long n, int r, int p) {
    if (n < r) return 0;
    if (r == 0) return 1;
    return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
           p;
  }
  int main() {
    c_p_c();
    int x = 1;
    cin >> x;
    while (x--) {
      long long n, k;
      cin >> n >> k;
      long long arr[n];
      long long s = 0;
      for (long long i = 0; i < n; i++) {
        cin >> arr[i];
        s += arr[i];
      }
      sort(arr, arr + n);
      long long x = arr[0];
      if (arr[0] * n <= k) {
        long long i = n - 1;
        long long ans = 0;
        while (s > k && i > 0) {
          s -= arr[i] - arr[0];
          ans++;
          i--;
        }
        long long ans2 = ans;
        i++;
        while (i < n) {
          if (s + arr[i] - arr[0] <= k) {
            ans2--;
            ans = min(ans, ans2);
            s += arr[i] - arr[0];
          } else {
            long long y2 = s - k + arr[i] - arr[0];
            long long y3;
            if (y2 % (n - i) == 0)
              y3 = y2 / (n - i);
            else
              y3 = y2 / (n - i) + 1;
            ans2 += y3 - 1;
            s += arr[i] - arr[0];
            s -= (n - i) * (y3);
            arr[0] -= y3;
            ans = min(ans, ans2);
          }
          i++;
        }
        cout << ans << endl;
      } else {
        long long y = k / n;
        long long ans = arr[0] - y;
        s -= arr[0] - y;
        long long i = n - 1;
        arr[0] = y;
        while (s > k && i > 0) {
          s -= arr[i] - arr[0];
          ans++;
          i--;
        }
        long long ans2 = ans;
        i++;
        while (i < n) {
          if (s + arr[i] - arr[0] <= k) {
            ans2--;
            ans = min(ans, ans2);
            s += arr[i] - arr[0];
          } else {
            long long y2 = s - k + arr[i] - arr[0];
            long long y3;
            if (y2 % (n - i) == 0)
              y3 = y2 / (n - i);
            else
              y3 = y2 / (n - i) + 1;
            ans2 += y3 - 1;
            s += arr[i] - arr[0];
            s -= (n - i) * (y3);
            arr[0] -= y3;
            ans = min(ans, ans2);
          }
          i++;
        }
        cout << ans << endl;
      }
    }
    return 0;
  }
  