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.*;
  public class Codeforces {
  
  
  	// CHECK CONSTRAINTS ALWAYS EDGE CASE MISSED 
  	
  static int mod = 1000000007;	
  
  
  
  
  public static void main(String[] args) {
       Scanner sc = new Scanner(System.in);
          while(sc.hasNext()) {
     int t   =  sc.nextInt();
     while(t-->0) {
  	  	 
  	int n =  sc.nextInt();
  	long k = sc.nextLong();
  	long a[] = new long[n+1];
  	for(int i = 1 ; i<=n;i++) {
  		a[i] = sc.nextLong();
  	}
  	a[0] = -1;
  	Arrays.sort(a);
  	
  	long sum = 0;
  	for(int i = 1 ; i<=n;i++) {
  		sum+=a[i];
  	}
  	
  	if(sum<=k) {
  		System.out.println(0);
  	}
  	else if(sum == k+1) {
  		System.out.println(1);
  	}
  	else {
  		long ans = sum-k;
  		
  		long diff = sum - k;
  	         sum = 0;
  	  for(int i = n ; i>1 ; i--) {
  		long x = 0;
  		  sum+=a[i];
  		  if(sum - (n-i+1)*a[1] >= diff) {
  			  x = 0;
  		  }
  		  else {
  			  x = (long) Math.max(Math.ceil((double)(diff - sum + ((n-i+1)*a[1]))/(n-i+2)) , 0);
  	      
  		  }
  		
  		  ans = Math.min(ans, x +n-i+1);
  		  
  	  }
  	  System.out.println(ans);
  	         
  	         
  	}
  	
  	
  	
  	
  
  		
  	 
  	   
  	   
  	   
  	   
  	   
  	   
  	   
     }
       }
      }
      }
  
          
  
  
  
  
  		
  
  
          
  
  
  
  
  		