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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Iterator;
import java.util.StringTokenizer;

public class AndItsNonZero {
	
	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 nextFloat() { return Float.parseFloat(next()); }
 
        double nextDouble() { return Double.parseDouble(next()); }
 
        String nextLine()
        {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
        int[] nextArray(int n)
        {
        	int[] a = new int[n];
        	for (int i=0; i<n; i++) {
        		a[i] = this.nextInt();
        	}
        	return a;
        }
        
        long[] nextArrayLong(int n)
        {
        	long[] a = new long[n];
        	for (int i=0; i<n; i++) {
        		a[i] = this.nextLong();
        	}
        	return a;
        }
        
    }
	
	public static void solve(int l, int r, int[][] a) {
		
		int max = 0;
		for (int i=0; i<20; i++) {
			max = Math.max(max, a[i][r]-a[i][l-1]);
		}
		System.out.println(r-l-max+1);
		
	}

	public static void main(String[] args) {
		
		FastReader in = new FastReader();
		int[][] a = new int[20][200001];
		for (int i=0; i<20; i++) {
			int count = 0;
			int mask = 1 << i;
			for (int j=1; j<=200000; j++) {
				if ((j & mask) != 0) {
					count++;
				}
				a[i][j] = count;
			}
		}
		int t = in.nextInt();
		while (t-- > 0) {
			solve(in.nextInt(), in.nextInt(), a);
		}

	}

}
