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.InputStreamReader;
import java.io.BufferedReader;
public class HelloWorld {
    public static void main(String []args) throws java.io.IOException {
        BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(scan.readLine());
        while(t-- > 0) {
            String[] lr = scan.readLine().split(" ");
            int l = Integer.parseInt(lr[0]);
            int r = Integer.parseInt(lr[1]);
            int start = 0, end = 0;
            for(int i=0; i<31; i++) start = Math.max(start, l & (1 << i));
            for(int i=0; i<31; i++) end = Math.max(end, r & (1 << i));
            if(start == end) {
            	System.out.println(0);
            }
            else if(end/start == 2){
            	int temp = Math.min(start, r - end + 1);
            	temp -= (l-start); 
            	System.out.println(temp);	
            }
            else {
            	int powStart = power(start);
            	int powEnd = power(end);
            	int noOfEle = powEnd - powStart - 1;
            	int sum = start * ((int)Math.pow(2.0, noOfEle * 1.0) - 1);
            	int temp = Math.min(start, r - end + 1);
            	temp -= (l-start); 
            	temp += sum;
            	System.out.println(temp);
            }
        }
    }
    private static int power(int num) {
    	int count = 0;
    	while(num != 1) {
    		num /= 2;
    		count++;
    	}
    	return count;
    }
}