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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
  //import sun.nio.fs.RegistryFileTypeDetector;
  
  
   
  public class B {
  	static int mod= 1000000007;
  	
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		
  		outer:while(t-->0) {
  			int n=fs.nextInt();
  			long k=fs.nextLong();
  			int arr[]=fs.readArray(n);
  			sort(arr);
  			long brr[]=new long[n];
  			for(int i=0;i<n;i++) brr[i]=arr[i];
  			long sum=0;
  			for(int ele:arr) sum+=ele;
  			if(sum<=k) {
  				out.println(0);
  				continue;
  			}
  			
  			long l=arr[0]-Math.abs(k-sum) , r=arr[0];
  			
  //			l=r;
  			
  			while(r-l>100) {
  //				System.out.println("Hree");
  				long m1= l+ (r-l)/3;
  				long m2=r-(r-l)/3;
  				long f1= find(brr,k,m1);
  				long f2=find(brr,k,m2);
  				if(f1<f2) {
  					l=m1;
  				}
  				else {
  					r=m2;
  				}
  			}
  			long ans=Long.MIN_VALUE;
  			
  			long lol=0;
  			for(long i=l;i<=r;i++) {
  				long cur=find(brr,k,i);
  				if(cur>ans) {
  					ans=cur;
  					lol=i;
  				}
  				
  				
  			}
  //			for(long i=-100;i<=arr[0];i++) {
  //				long cur=-find(brr,k,i);
  //				out.println(cur+" "+i);
  //			}
  			
  //			System.out.println(lol);
  //			for(int i=-100;i<=1;i++) {
  //				System.out.println(find(brr,k,i));
  //			}
  			out.println(-ans);
  			
  			
  			
  		}
  		
  		out.close();
  		
  	}
  	static long find(long arr[], long k,long to) {
  //		System.out.println("Herer");
  		int n=arr.length;
  		long ans=0;
  		ans+=arr[0]-to;
  //		arr[0]=to;
  		long sum=0;
  		sum+=to-arr[0];
  		for(long ele:arr) sum+=ele;
  		int ind=n-1;
  		while(sum>k&&ind>0) {
  			if(arr[ind]==to) {
  				break;
  			}
  			sum-=arr[ind]-to;
  //			arr[ind]=arr[0];
  			ans++;
  			ind--;
  		}
  		if(sum<=k) {
  			return -ans;
  		}
  		
  		ans+=sum-k;
  		return -ans;
  	}
  	static long pow(long a,long b) {
  		if(b<0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static int gcd(int  a,int  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	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[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		int[] readArray(int n) {
  			int[] a=new int[n];
  			for (int i=0; i<n; i++) a[i]=nextInt();
  			return a;
  		}
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  	}
   
  }