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: //package setordecrease;
  
  import java.util.*;
  import java.io.*;
  
  public class setordecrease {
  	
  	public static boolean isValid(long val, Long[] nums, long k) {
  		long sum = 0;
  		long min = nums[nums.length - 1];
  		for(int i = 0; i < nums.length - 1; i++) {
  			sum += nums[i];
  		}
  		if(val > nums.length - 1) {
  			min -= (val - (nums.length - 1));
  			val = nums.length - 1;
  		}
  		for(int i = 0; i < val; i++) {
  			sum -= (nums[i] - min);
  		}
  		if(sum + min <= k) {
  			return true;
  		}
  		for(int i = (int) val - 1; i >= 0; i--) {
  			sum += nums[i] - min;
  			sum -= i;
  			min --;
  			if(sum + min <= k) {
  				return true;
  			}
  		}
  		return false;
  	}
  	
  	public static void main(String[] args) throws IOException {
  		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
  		int t = Integer.parseInt(fin.readLine());
  		StringBuilder fout = new StringBuilder();
  		while(t-- > 0) {
  			StringTokenizer st = new StringTokenizer(fin.readLine());
  			int n = Integer.parseInt(st.nextToken());
  			long k = Long.parseLong(st.nextToken());
  			st = new StringTokenizer(fin.readLine());
  			Long[] nums = new Long[n];
  			for(int i = 0; i < n; i++) {
  				nums[i] = Long.parseLong(st.nextToken());
  			}
  			Arrays.sort(nums, (a, b) -> -Long.compare(a, b));
  			long ans = Long.MAX_VALUE / 2;
  			long high = ans;
  			long low = 0;
  			long mid = low + (high - low) / 2;
  			while(low <= high) {
  				if(isValid(mid, nums, k)) {
  					ans = Math.min(ans, mid);
  					high = mid - 1;
  				}
  				else {
  					low = mid + 1;
  				}
  				mid = low + (high - low) / 2;
  			}
  			fout.append(ans).append("\n");
  		}
  		System.out.print(fout);
  	}
  }
  