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.
/* package codechef; // don't place package name! */

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

public class globalRoundC {
    static int mod = (int) (1e9 + 7);
    static long count = 0;

    public static void main(String[] args) throws java.lang.Exception {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();

        int[][] dp = new int[200001][33];
        for (int i = 1; i <= 200000; i++) {
            for (int j = 0; j < 33; j++) {
                int m = 1 << j;
                if ((i & m) != 0)
                    dp[i][j] = dp[i - 1][j] + 1;
                else
                    dp[i][j] = dp[i - 1][j];
                // dp[i][j]
            }
        }
        // printmat(dp);

        while (t-- > 0) {

            int l = s.nextInt();
            int r = s.nextInt();

            int n = r - l + 1;
            // int[] arr = new int[2*100000 + 1];
            // int l = scn.nextInt();
            // int r = scn.nextInt();
            int max1 = 0;

            for (int j = 0; j <= 32; j++) {
                // int currmax= 0;
                // currmax = Math.abs(dp[l][j] - dp[r][j]);
                // System.out.println(dp[l][j] + " " + d p [l-1][j] +  " "+ dp[r][j]);
                max1 = Math.max(dp[r][j] - dp[l - 1][j], max1);
            }
            // max1= Math.max(currmax,max1);
            // }
            int del = n - max1;
            System.out.println(del);

            // long maxsetbit = 0;
            // for (int i = 0 ; i< arr.length; i++){
            // maxsetbit = Math.max(maxsetbit, arr[i]);
            // }

        }
    }

    public static void printmat(int[][] matrix) {
        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                System.out.print(matrix[i][j] + " ");
            }
            System.out.println();
        }
    }

}