You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?

A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.

The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.

Output

For each test case, output a single integer — the answer to the problem.

Example

Input


5
1 2
2 8
4 5
1 5
100000 200000


Output


1
3
0
2
31072

Note

In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.

In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.

import java.io.*;
import java.util.*;
public class solution {
	static long cr[][]=new long[1001][1001];
   static long mod=1000000007;
public static void main(String[] args)  {
	FScanner sc = new FScanner();
	//Arrays.fill(prime, true);
	//sieve();
	
	//ncr();
	int t=sc.nextInt();
	 StringBuilder sb = new StringBuilder();
	while(t-->0) {	
     int l=sc.nextInt();
     int r=sc.nextInt();
     long res=0;
     ArrayList<Integer> A=new ArrayList<>();
     for(int i=0;i<32;i++)
     {
    	 if(Math.pow(2, i)>=l && Math.pow(2, i)<=r)
    	 {
    		 A.add(i);
    	 }
     }
     
     if(A.size()==1)
     {
    	 res=Math.max( (int)Math.pow(2,A.get(0))-l ,r- (int)Math.pow( 2,A.get(0))+1);
    	 res=r-l+1-res;
     }
     else if(A.size()>1)
     {
    	 res=(int)Math.pow(2,A.get(A.size()-1)) - (int)Math.pow(2,A.get(A.size()-2));
    	 res=Math.max(r-(int)Math.pow(2,A.get(A.size()-1)), res);
    	 res=r-l+1-res;
     }
     if(l%2==1 && r%2==1)
     res=Math.min( r-l -(r-l+1)/2 , res);
     else
    res=Math.min( (r-l+1)/2 , res) ;
    	sb.append(res); 
     
       sb.append("\n");			
	}
	  System.out.println(sb.toString());	
}



public static long gcd(long a,long b)
{   
	
     return b==0 ? a:gcd(b,a%b);	
	
}

        
 /* public static void sieve()
{   prime[0]=prime[1]=false;
int n=1000000;
	for(int p = 2; p*p <=n; p++)
    {
        if(prime[p] == true)
        {
          
            for(int i = p*p; i <= n; i += p)
                prime[i] = false;
        }
    }
   */

public static void ncr()
{
	cr[0][0]=1;
	for(int i=1;i<=1000;i++)
	{
		cr[i][0]=1;
		for(int j=1;j<i;j++)
		{
			cr[i][j]=(cr[i-1][j-1]+cr[i-1][j])%mod;
		}
		cr[i][i]=1;
	}
}

}
 
  class pair implements Comparable<pair> {
	int val;int ind;
	 pair(int val,int ind)
	{
		this.val=val;
		this.ind=ind;
	}
	 public int compareTo(pair p)
	 {
		 return p.ind-this.ind;
	 }
	
	
}  
 
 class FScanner {
	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	StringTokenizer sb = new StringTokenizer("");
 
	String next(){
		while(!sb.hasMoreTokens()){
			try{
				sb = new StringTokenizer(br.readLine());
			} catch(IOException e){ }
		}
		return sb.nextToken();
	}
	String nextLine(){
		try{ return br.readLine(); } 
		catch(IOException e) { } return "";
	}
 
	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;
	}
 
	float nextFloat(){
	return Float.parseFloat(next());
	}
	
	double nextDouble(){
	return Double.parseDouble(next());
	}
}