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: /*
          "Everything in the universe is balanced. Every disappointment
                  you face in life will be balanced by something good for you!
                          Keep going, never give up."
  
  */
  
  
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Solution {
      static long MAX = 1000000000000000000L;
      public static void main(String[] args) throws java.lang.Exception {
          out = new PrintWriter(new BufferedOutputStream(System.out));
          sc = new FastReader();
  
          int test = sc.nextInt();
          for (int t = 0; t < test; t++) {
              solve();
          }
          out.close();
      }
  
      private static void solve() {
          int n = sc.nextInt();
          long k = sc.nextLong();
          int[] arr = new int[n];
          for (int i = 0; i < n; i++) {
              arr[i] = sc.nextInt();
          }
          sort(arr);
          long[] pref = new long[n + 1];
          for (int i = 0; i < n; i++) {
              pref[i + 1] = pref[i] + arr[i];
          }
          long minOperations = MAX;
          for (int i = 0; i <= n - 1; i++) { // performing operation of type 2 on last i elements of the array (changing last i elements to min value arr[0]).
              long sum = pref[n - i] + (long) arr[0] * i;
              long operations = i;
              if (sum > k) { // need more operations
                  long difference = sum - k;
                  operations += (difference + i) / (i + 1);
              }
              minOperations = Math.min(minOperations, operations);
          }
          out.println(minOperations);
      }
  
      private static void sort(int[] arr) {
          List<Integer> list = new ArrayList<>();
          for (int val : arr) {
              list.add(val);
          }
          Collections.sort(list);
          int n = list.size();
          for (int i = 0; i < n; i++) {
              arr[i] = list.get(i);
          }
      }
  
  
      public static FastReader sc;
      public static PrintWriter out;
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer str;
  
          public FastReader()
          {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next()
          {
              while (str == null || !str.hasMoreElements())
              {
                  try
                  {
                      str = new StringTokenizer(br.readLine());
                  }
                  catch (IOException  end)
                  {
                      end.printStackTrace();
                  }
              }
              return str.nextToken();
          }
  
          int nextInt()
          {
              return Integer.parseInt(next());
          }
  
          long nextLong()
          {
              return Long.parseLong(next());
          }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try
              {
                  str = br.readLine();
              }
              catch (IOException end)
              {
                  end.printStackTrace();
              }
              return str;
          }
      }
  }