You are given two integers l and r, l≤ r. Find the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

As a reminder, a mod b is a remainder we get when dividing a by b. For example, 26 mod 8 = 2.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1≤ t≤ 10^4), denoting the number of test cases. Description of the test cases follows.

The only line of each test case contains two integers l, r (1≤ l ≤ r ≤ 10^9).

Output

For every test case, output the largest possible value of a mod b over all pairs (a, b) of integers for which r≥ a ≥ b ≥ l.

Example

Input


4
1 1
999999999 1000000000
8 26
1 999999999


Output


0
1
12
499999999

Note

In the first test case, the only allowed pair is (a, b) = (1, 1), for which a mod b = 1 mod 1 = 0.

In the second test case, the optimal choice is pair (a, b) = (1000000000, 999999999), for which a mod b = 1.
import java.util.Scanner;

public class A {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t=sc.nextInt();
		while(t-->0) {
			long l=sc.nextInt();
			long r=sc.nextInt();
			
//			boolean flag=false;
//			for(long i=r;i>=l;i--) {
//				if((i*2)<=r) {
//					flag=true;
////					System.out.println(i);
//					System.out.println(r%(i+1));
//					break;
//				}
//			}
//			
//			if(flag==false) {
//				System.out.println(r%l);
//			}
			
			long mid=(r/2);
			
			if(r%2==0) {
				if(mid<l) {
					System.out.println(r%l);
				}
				else {
					System.out.println(r%(mid+1));
				}
			}
			else {
				if(mid<l) {
					System.out.println(r%l);
				}
				else {
					System.out.println(r%(mid+1));
				}
			}
			
			
		}
	}
}
