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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  
  public class sets_or_decrease {
  	
  	public static long ceil_div(long a, long b) {return a % b == 0 ? a / b : a / b + 1;}
  	public static void sort(int[] arr)
      {
          ArrayList<Integer> ls = new ArrayList<Integer>();
          for(int x: arr)
              ls.add(x);
          Collections.sort(ls);
          for(int i=0; i < arr.length; i++)
              arr[i] = ls.get(i);
      }
  	public static void main (String[] args) throws IOException
  	{
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); 
  	
  		int t = Integer.parseInt(br.readLine());
  		for(int f=0;f<t;f++) {
  			StringTokenizer st = new StringTokenizer(br.readLine());
              int n = Integer.parseInt(st.nextToken());
              long k = Long.parseLong(st.nextToken());
  			int[] arr0= Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
  			sort(arr0);
  			
  			long[] arr=new long [n+1];
  			for(int i=1;i<n+1;i++) {
  				arr[i]=arr0[i-1];
  			}
  			long sumGlobal=0;
  			for(int j=1;j<arr.length;j++) {
  				sumGlobal+=arr[j];
  			}
  			double result=sumGlobal-k;
  			
  			long diff = sumGlobal - k;
  			long sum = 0;
  			
  			for (int i = n; i > 1; i--)
  			{
  				sum += arr[i];
  				long x;
  				if (sum - (n - i + 1)*arr[1] >= diff)x = 0;
  				else x = Math.max(ceil_div(diff - sum + (n - i + 1) * arr[1], n - i + 2), 0L);
  				result = Math.min(result, x + n - i + 1);
  			}
  			
  			if(n==1) {
  				result=Math.abs(arr[1]-k);
  			}
  			if(sumGlobal<=k) {
  				result=0;
  			}
  			System.out.println((int)result);
  		}
  		
  	}
  }