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.util.*;
  import java.io.*;
  
  public class Codeforces {
  
    static long accurateFloor(long a, long b) {
      long val = a/b;
      while(val*b > a) val--;
      return val;
    }
  
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        int T = Integer.parseInt(br.readLine());
        while(T-- > 0) {
          StringTokenizer st = new StringTokenizer(br.readLine());
          int n = Integer.parseInt(st.nextToken());
          long k = Long.parseLong(st.nextToken());
          Long[] a = new Long[n];
          st = new StringTokenizer(br.readLine());
          long[] sum = new long[n];
          for(int i = 0; i < n; i++) {
            a[i] = Long.parseLong(st.nextToken());
          }
          Arrays.sort(a);
          sum[0] = a[0];
          for(int i = 1; i < n; i++) {
            sum[i] += sum[i-1]+a[i];
          }
          long ans = Long.MAX_VALUE;
          for(int y = 0; y < n; y++) {
            // long z = k-sum[n-y-1]+a[0];
            // if(z > 0) z = z/(y+1);
            // else {
            //   if(z%(y+1) == 0) z = z/(y+1);
            //   else {
            //     z = z/(y+1);
            //     z -= 1;
            //   }
            // }
            long x = a[0]-accurateFloor(k-sum[n-y-1]+a[0], y+1);
            x = Math.max(0, x);
            ans = Math.min(ans, y+x);
          }
          writer.println(ans);
        }
        writer.close();
        br.close();
    }
  
  }
  