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.lang.*;
  import java.io.*;
  import java.util.*;
  
  public class C {
     static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              try {
                  br = new BufferedReader(
                          new FileReader("input.txt"));
                  PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                  System.setOut(out);
              } catch (Exception e) {
                  br = new BufferedReader(new InputStreamReader(System.in));
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.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 e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      // end of fast i/o code
      public static void main(String[] args) {
          FastReader reader = new FastReader();
          int t = reader.nextInt();
          while(t-->0){
              int n = reader.nextInt();
              long k = reader.nextLong();
              long arr[] = new long[n+1];
              long totalSum = 0l, endSum = 0l;
              for (int indx = 1; indx<n+1 ; indx++) {
                  arr[indx] = reader.nextLong();
                  totalSum += arr[indx]; 
              }
              Arrays.sort(arr);
              long diff = totalSum - k;
              long minOps = diff;
              for(int indx = n; indx>1; indx--){
                  endSum += arr[indx];
                long x;
                  if (endSum - (n - indx + 1)*arr[1] >= diff)x = 0;
                  else x = Math.max(ceil_div(diff - endSum + (n - indx + 1) * arr[1], n - indx + 2), 0L);
                  minOps = Math.min(minOps, x + n - indx + 1);
              }
              if(n==1) {
                  minOps=Math.abs(arr[1]-k);
              }
              if(totalSum<=k) {
                  minOps=0;
              }
              System.out.println((int)minOps);
          }
      }
      private static long ceil_div(long a, long b) {return a % b == 0 ? a / b : a / b + 1;}
  }