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) {
  		FastScanner fs=new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int T = fs.nextInt();
          for (int tt=0; tt<T; tt++) {
  			int n = fs.nextInt();
  			long k = fs.nextLong();
              int[] a = fs.readArray(n);
  			ruffleSort(a);
  			long[] prefixSum = new long[n];
  			long sum = 0;
  			for (int i=0; i<n; i++) {
  				sum+=a[i];
  				prefixSum[i] = sum;
  			}
  			if (n == 1) {
  				out.println(Math.max(0, a[0]-k));
  				continue;
  			}
  			if (prefixSum[n-1]<=k) {
  				out.println(0);
  				continue;
  			}
  			long res = Long.MAX_VALUE;
  			for (int y = 0; y<n; y++) {
  				long psum = prefixSum[n-y-1];
  //				psum+(y+1)(a[0]-x)-a[0]=k
  //				x = a[0]-(k-psum+a[0])/(y+1)
  				long x = a[0]-(k-psum+a[0])/(y+1);
  				if (k-psum+a[0]<0 && (k-psum+a[0])%(y+1)!=0)
  					x++;
  				if (x>=0) {
  					res=Math.min(res, x+y);
  				} else {
  					res=Math.min(res, y);
  				}
  			}
  			out.println(res);
          }
          out.close();
  	}
  
  	static final Random random=new Random();
   
      static void ruffleSort(int[] a) {
          int n=a.length;//shuffle, then sort
          for (int i=0; i<n; i++) {
              int oi=random.nextInt(n), temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
      
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		String next() {
  			while (!st.hasMoreTokens()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
  
  	}
  }
  