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: import java.io.OutputStream;
  import java.io.IOException;
  import java.io.InputStream;
  import java.io.PrintWriter;
  import java.util.Scanner;
  import java.util.Collections;
  import java.util.ArrayList;
  
  /**
   * Built using CHelper plug-in
   * Actual solution is at the top
   */
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          Scanner in = new Scanner(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          TaskC solver = new TaskC();
          solver.solve(1, in, out);
          out.close();
      }
  
      static class TaskC {
          public void solve(int testNumber, Scanner in, PrintWriter out) {
              int T = in.nextInt();
              while (T-- > 0) {
                  solveOne(in, out);
              }
          }
  
          private void solveOne(Scanner in, PrintWriter out) {
              int N = in.nextInt();
              long K = in.nextLong();
              long A[] = L.readLongArray(N, in);
              L.sort(A);
              long[] prefSum = new long[N + 1];
              for (int i = 0; i < N; i++) {
                  prefSum[i + 1] = prefSum[i] + A[i];
              }
              long ans = Long.MAX_VALUE;
              for (int t = 0; t <= N - 1; t++) {
                  long sum = prefSum[N - t] + A[0] * t;
                  long steps = t;
                  if (sum > K) {
                      long diff = sum - K;
                      steps += Math.ceil((double) diff / (t + 1));
                  }
                  ans = Math.min(ans, steps);
              }
              out.println(ans);
          }
  
      }
  
      static class L {
          public static long[] readLongArray(int size, Scanner in) {
              long[] array = new long[size];
              for (int i = 0; i < size; i++) {
                  array[i] = in.nextLong();
              }
              return array;
          }
  
          public static void sort(long[] a) {
              ArrayList<Long> l = new ArrayList<>();
              for (long i : a) l.add(i);
              Collections.sort(l);
              for (int i = 0; i < a.length; i++) a[i] = l.get(i);
          }
  
      }
  }
  
  