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.PrintWriter;
import java.util.Scanner;

public class Main {
    
    public static void main(String[] args) {
        
        var sc = new Scanner(System.in);
        var pw = new PrintWriter(System.out);
        
        var bit = new BinaryIndexedTree[18];
        for(int i = 0; i < 18; i++){
            bit[i] = new BinaryIndexedTree(200000);
        }
        for(int i = 1; i <= 200000; i++){
            String s = Integer.toBinaryString(i);
            int len = s.length();
            for(int j = 0; j < len; j++){
                if(s.charAt(j) == '1'){
                    bit[len - 1 - j].add(i, 1);
                }
            }
        }
        
        int T = Integer.parseInt(sc.next());
        for(int t = 0; t < T; t++){
            int l = Integer.parseInt(sc.next());
            int r = Integer.parseInt(sc.next());
            
            long max = 0;
            for(int i = 0; i < 18; i++){
                max = Math.max(bit[i].sum(l, r+1), max);
            }
            pw.println(r-l+1-max);
        }
        pw.flush();
    }
    
    static class BinaryIndexedTree {
        int n;
        long[] bit;
        
        public BinaryIndexedTree(int n){
            this.n = n;
            bit = new long[n+1];
        }
        void add(int i, int x){
            while(i <= n){
                bit[i] += x;
                i += i & -i;
            }
        }
        long sum(int l, int r){
            return sum(r-1) - sum(l-1);
        }
        long sum(int i){
            long s = 0;
            while(i > 0){
                s += bit[i];
                i -= i & -i;
            }
            return s;
        }
    }
}