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.time.Year;
import java.util.Scanner;
import java.util.*;
public class test2 {


    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
//            测试次数
        int testnum = in.nextInt();
        for (int k = 0; k <testnum; k++) {
            int a = in.nextInt();
            int b = in.nextInt();
            int c = baoli(a, b);
                    int[] ansa = check(1, a);
                    int[] ansb = check(1, b);
                    int min = Integer.MAX_VALUE;
                for (int i = 0; i < ansb.length; i++) {
                    int num = 1 << i;
                    ansb[i] += (num & a) == 0 ? 1 : 0;
                }
                    for (int i = 0; i <ansa.length; i++) {

                        min = Math.min(min, ansb[i] - ansa[i]);
                    }
                    for (int i = ansa.length; i < ansb.length; i++) {
                        min = Math.min(min, (1<<(ansb.length-1))-a);
                    }
            System.out.println(c+"---"+min);



        }
    }
    public static int[] check(int a,int b){
        String sa = Integer.toBinaryString(a);
        String sb = Integer.toBinaryString(b);
        int n = sb.length();
        int m = sa.length();
        int[] ans = new int[n];
        int min = Integer.MAX_VALUE;
        int pre = 0;
//            0-> 1<<n 0的个数正好是n-1
        int sumb = 0;
        for (int i = n-1; i>=0; i--) {
            int num=1<<i;
            sumb += num;
            int flag = sb.charAt(n-i-1) - '0';
            if(flag==0){
                sumb -= num;
                ans[i] = pre + b - sumb +1;
            }else{
                ans[i] = pre + num ;
                pre += num/2;
            }
//            System.out.print(ans[i]+"~~~");
        }

        return ans;
    }
    static int baoli(int a,int b){
        int[] ans = new int[31];
        for (int i = a; i <=b ; i++) {
            for (int j = 0; j < 31; j++) {
                int num = 1 << j;
//                System.out.print((num & i)+"~~~~~");
                ans[j] += (num & i)==0? 1:0;
            }
        }
//        System.out.println();
        int min = Integer.MAX_VALUE;
        for (int i = 0; i <31 ; i++) {
//            System.out.print(ans[i]+"~~~~~");
            min = Math.min(ans[i], min);
        }
        return min;
    }



}
