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.*;
  import java.io.*;
  
  ////***************************************************************************
   /* public class E_Gardener_and_Tree implements Runnable{
  
         public static void main(String[] args) throws Exception {
          new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
         }
      public void run(){
           WRITE YOUR CODE HERE!!!!
           JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
         }
  
    }
  */
  /////**************************************************************************
  
  
  public class C_Set_or_Decrease{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          //PrintWriter out=new PrintWriter(System.out);
          //end of program
          //out.println(answer);
          //out.close();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();
          long k=s.nextLong();
          ArrayList<Long> list = new  ArrayList<Long>();
          long sum=0;
          for(int i=0;i<n;i++){
              long a=s.nextLong();
              list.add(a);
              sum+=a;
          }
          if(sum<=k){
              res.append("0 \n");
          }
          else{
              Collections.sort(list);
              long well[]= new long[n-1];
              long ss=0;
              for(int i=1;i<n;i++){
                  long a=list.get(i);
                  ss+=a;
                  well[i-1]=ss;
                //  System.out.print(well[i-1]+" ");
              }
              System.out.println();
              long yo=k/n;
              long first=list.get(0);
              long yo1=Math.min(yo,first);
              long ans=Long.MAX_VALUE;
             // System.out.println("yo1 "+yo1);
              for(long i=yo1;i>(yo1-n-5);i--){
                //  System.out.println("i "+i);
                  long num=first-i;
                  long start=-1;
                  long end=n-1;
                  long mid=0;
                  while(start<end){
                      mid=(start+end)/2;
                      if(start+1==end){
                          break;
                      }
                      long hh=well[(int)mid];
                      long a1=(n-mid-1);
                      long a2=(a1*i);
                      hh+=a2;
                      if(hh<=k){
                          start=mid;
                      }
                      else{
                          end=mid;
                      }
                  }
                  long jj=n-start-2;
                  long kk=jj+num;
                  ans=Math.min(ans,kk);
  
  
  
              }
              res.append(ans+" \n");
          }
          p++;
      }
      System.out.println(res);
  
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
      static long modpower(long x, long y, long p)
       {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
  
    
     // SIMPLE POWER FUNCTION=>
  
  
        static long power(long x, long y)
          {
              long res = 1;     // Initialize result
          
              while (y > 0)
              {
              
                  // If y is odd, multiply x with result
                  if ((y & 1) != 0)
                      res = res * x;
          
                  // y must be even now
                  y = y >> 1; // y = y/2
                  x = x * x;  // Change x to x^2
              }
              return res;
          }
  
  
  
  
  
  }