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.*;
  import java.util.*;
  
  public class C {
  	public static void main (String[] args) throws IOException {
  		Kattio io = new Kattio();
  		int t = io.nextInt();
  		for (int iii=0; iii<t; iii++) {
  			int n = io.nextInt();
  			long k = io.nextLong();
  
  			Long[] arr = new Long[n];
  			for (int i=0; i<n; i++) {
  				arr[i] = io.nextLong();
  			}
  
  			Arrays.sort(arr);
  			long[] psum = new long[n];
  			psum[0] = arr[0];
  			for (int i=1; i<n; i++) {
  				psum[i] = psum[i-1] + arr[i];
  			}
  
  			if (psum[n-1] <= k) {
  				System.out.println(0);
  				continue;
  			}
  			//System.out.println(psum[n-1]);
  
  			long ans = Long.MAX_VALUE;
  			
  			for (int i=0; i<n; i++) {
  				long sum = psum[i] + (n-i-1)*arr[0];
  				if (sum <= k) {
  					ans = Math.min(ans, n-1-i);
  					continue;
  				}
  				long ops = (long) Math.ceil((double) (sum - k)/(double) (n-i));
  
  				ans = Math.min(ans, n-1-i + ops);
  				//System.out.println(sum);
  			}
  			System.out.println(ans);
  			
  		}
  	}
  
  	static class Kattio extends PrintWriter {
  		private BufferedReader r;
  		private StringTokenizer st;
  
  		// standard input
  		public Kattio() { this(System.in, System.out); }
  		public Kattio(InputStream i, OutputStream o) {
  			super(o);
  			r = new BufferedReader(new InputStreamReader(i));
  		}
  		// USACO-style file input
  		public Kattio(String problemName) throws IOException {
  			super(new FileWriter(problemName + ".out"));
  			r = new BufferedReader(new FileReader(problemName + ".in"));
  		}
  
  		// returns null if no more input
  		public String next() {
  			try {
  				while (st == null || !st.hasMoreTokens())
  					st = new StringTokenizer(r.readLine());
  				return st.nextToken();
  			} catch (Exception e) { }
  			return null;
  		}
  
  		public int nextInt() { return Integer.parseInt(next()); }
  		public double nextDouble() { return Double.parseDouble(next()); }
  		public long nextLong() { return Long.parseLong(next()); }
  	}
  }