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;
  vector<long long> sieve(long long n) {
    long long* arr = new long long[n + 1]();
    vector<long long> vect;
    for (long long i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (long long j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  long long expo(long long a, long long b, long long MOD) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % MOD;
      a = (a * a) % MOD;
      b = b >> 1;
    }
    return res;
  }
  long long gcd(long long a, long long b) {
    if (a == 0) return b;
    return gcd(b % a, a);
  }
  long long findGCD(long long arr[], long long n) {
    long long result = arr[0];
    for (long long i = 1; i < n; i++) {
      result = gcd(arr[i], result);
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  string decToBinary(long long n) {
    string output = "";
    while (n) {
      if (n & 1) {
        output = '1' + output;
      } else {
        output = '0' + output;
      }
      n >>= 1;
    }
    return output;
  }
  long long isSubstring(string s1, string s2) {
    long long M = s1.length();
    long long N = s2.length();
    for (long long i = 0; i <= N - M; i++) {
      long long j;
      for (j = 0; j < M; j++)
        if (s2[i + j] != s1[j]) break;
      if (j == M) return i;
    }
    return -1;
  }
  void solve() {
    long long n, k;
    cin >> n >> k;
    long long arr[n];
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    sort(arr, arr + n);
    long long sum[n + 1];
    sum[1] = arr[0];
    for (long long i = 1; i < n; i++) {
      sum[i + 1] = sum[i] + arr[i];
    }
    long long ans = 1e18;
    for (long long i = 0; i < n; i++) {
      long long x;
      long long m = k + arr[0] - sum[n - i];
      if ((m / (i + 1)) * (i + 1) > m) {
        x = arr[0] - (m / (i + 1) - 1);
      } else {
        x = arr[0] - (m / (i + 1));
      }
      ans = min(ans, i + max(0ll, x));
    }
    cout << ans << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  