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;
  template <class T>
  void _p(T a) {
    cout << a << '\n';
  }
  template <class T>
  void _pa(T a) {
    cout << a << ' ';
  }
  template <class T1, class T2>
  void _p(T1 a, T2 b) {
    cout << a << ' ' << b << '\n';
  }
  template <class T1, class T2, class T3>
  void _p(T1 a, T2 b, T3 c) {
    cout << a << ' ' << b << ' ' << c << '\n';
  }
  template <class T1, class T2, class T3, class T4>
  void _p(T1 a, T2 b, T3 c, T4 d) {
    cout << a << ' ' << b << ' ' << c << ' ' << d << ' ' << '\n';
  }
  template <class T1, class T2, class T3, class T4, class T5>
  void _p(T1 a, T2 b, T3 c, T4 d, T5 e) {
    cout << a << ' ' << b << ' ' << c << ' ' << d << ' ' << e << ' ' << '\n';
  }
  template <class T>
  void _pa(T a[], long long n) {
    long long i;
    for (i = 0; i < n; i++) cout << a[i] << ' ';
    cout << '\n';
  }
  template <class T>
  void _p(vector<T> v) {
    for (auto x : v) cout << x << ' ';
    cout << '\n';
  }
  template <class T>
  void _p(list<T> l) {
    for (auto x : l) cout << x << ' ';
    cout << '\n';
  }
  template <class T>
  void _p(set<T> s) {
    for (auto x : s) cout << x << ' ';
    cout << '\n';
  }
  template <class T>
  void _p(multiset<T> s) {
    for (auto x : s) cout << x << ' ';
    cout << '\n';
  }
  template <class T>
  void _p(unordered_set<T> s) {
    for (auto x : s) cout << x << ' ';
    cout << '\n';
  }
  template <class T1, class T2>
  void _p(map<T1, T2> m) {
    for (auto x : m) cout << x.first << "->" << x.second << '\n';
    cout << '\n';
  }
  template <class T1, class T2>
  void _p(unordered_map<T1, T2> m) {
    for (auto x : m) cout << x.first << "->" << x.second << '\n';
    cout << '\n';
  }
  template <class T1, class T2>
  void _p(map<T1, vector<T2>> m) {
    for (auto x : m) {
      cout << x.first << "->";
      _p(x.second);
    }
    cout << '\n';
  }
  template <class T1, class T2>
  void _p(unordered_map<T1, vector<T2>> m) {
    for (auto x : m) {
      cout << x.first << "->";
      _p(x.second);
    }
    cout << '\n';
  }
  void solution() {
    long long i, n, k, ans = LLONG_MAX, s, p, z, sm;
    cin >> n >> k;
    long long a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    for (i = 0; i < n; i++) s += a[i];
    sm = a[0] * n;
    vector<long long> v;
    for (i = 0; i < n; i++) {
      sm += (a[i] - a[0]);
      v.push_back(sm);
      if (sm <= k) ans = min(ans, n - 1 - i);
    }
    for (i = 0; i < n; i++)
      if (v[i] > k) {
        sm = v[i] - k;
        z = sm / (n - i) + (sm % (n - i) == 0 ? 0 : 1);
        ans = min(ans, n - 1 - i + z);
      }
    p = -1;
    sm = s;
    z = a[0] + 1;
    sm = s - a[0] - 1;
    if (sm <= k) ans = min(ans, z);
    for (i = 0; i < n - 1; i++) {
      sm -= a[n - 1 - i];
      sm--;
      z++;
      if (sm <= k) ans = min(ans, z);
    }
    p = 0;
    sm = s;
    z = a[0];
    sm = s - a[0];
    if (sm <= k) ans = min(ans, z);
    for (i = 0; i < n - 1; i++) {
      sm -= a[n - 1 - i];
      z++;
      if (sm <= k) ans = min(ans, z);
    }
    _p(ans);
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int tcse;
    cin >> tcse;
    while (tcse--) solution();
  }
  