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;
  const long long mod = 1e9 + 7;
  const long long maxn = 2e5 + 9;
  void init() {}
  vector<long long> preSum;
  bool solve(vector<long long>& arr, long long x, long long k, long long sum) {
    if (sum - x <= k) {
      return true;
    }
    long long tSum = sum;
    long long n = arr.size();
    if (x >= n - 1) {
      if (arr[n - 1] * n <= k) {
        return true;
      }
    }
    for (long long i = n - 2; i >= 0; --i) {
      long long remMoves = x - (i + 1);
      if (remMoves < 0) {
        continue;
      }
      long long smallestNum = arr[n - 1] - remMoves;
      long long newSum = tSum - preSum[i] - arr[n - 1] + smallestNum * (i + 2);
      if (newSum <= k) {
        return true;
      }
    }
    return false;
  }
  void test_case(long long tc) {
    long long n, k;
    cin >> n >> k;
    vector<long long> arr(n);
    long long sum = 0;
    for (long long i = 0; i <= n - 1; ++i) {
      cin >> arr[i];
      sum += arr[i];
    }
    preSum = vector<long long>(n, 0);
    sort(arr.begin(), arr.end(), greater<long long>());
    preSum[0] = arr[0];
    for (long long i = 1; i <= n - 1; ++i) preSum[i] = preSum[i - 1] + arr[i];
    long long start = 0, end = 1e18;
    long long ans = 0;
    while (end >= start) {
      long long mid = start + (end - start) / 2;
      if (solve(arr, mid, k, sum)) {
        ans = mid;
        end = mid - 1;
      } else {
        start = mid + 1;
      }
    }
    cout << ans << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
    long long t = 1;
    cin >> t;
    for (long long tc = 1; tc <= t; ++tc) test_case(tc);
  }
  