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.util.*;
import java.lang.Math;
import java.io.* ;


public class Account {
	
	public static class Pair<Object1 ,Object2> {
	    Object1 key ;
	    Object2 val ;
	    
	    Pair(Object1 key ,Object2 val) {
	        this.key = key ;
	        this.val = val ;
	    }
	    
	    Pair() {}
	    
	}
	
	public static void main (String[] args) {
		
//		Scanner sc = new Scanner(System.in) ;
		FastReader sc = new FastReader() ;
//		StringBuilder res = new StringBuilder() ;
		int n= 300000 ;
		int m= (int)(Math.log(n) / Math.log(2)) + 2 ;
		int[][] dp = new int[n+1][m] ;
		for(int i=1 ;i<dp.length ;i++) {
			for(int j=0 ;j<m ;j++) {
				 if((i & (1 << j)) != 0) {
					 dp[i][j] = dp[i-1][j] + 1 ;
				 }else {
					 dp[i][j] = dp[i-1][j] ;
				 }
			}
		}
		int t= 1 ;  
		t = sc.nextInt() ;
		while(t-- > 0) {
			int l= sc.nextInt() ;
			int r= sc.nextInt() ;
			int max= 0 ;
			for(int i=0 ;i<m ;i++) {
				max = Math.max(dp[r][i] - dp[l-1][i], max) ;
			}
			System.out.println((r-l+1-max));
  		}
	}
	
	
	/*
	 *   1--> a   3b   2c
	 *   2--> 2a  b    3c
	 *   3--> 3a  2b   c
	 */
		
	
	public static long gcd(long a, long b)
	{
	    if (b == 0)
	        return a;
	    return gcd(b, a % b);
	     
	}
	
	
//	_xor from 1 to n
	public static int xor(int n)
    {
        if (n % 4 == 0)
            return n;
      
        if (n % 4 == 1)
            return 1;
      
        if (n % 4 == 2)
            return n + 1;
      
        return 0;
    }

	
	public static class FastReader {
        
		BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                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());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
 
        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }	
	
}