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 Miracle_and_Sleeper {
	public static void main(String[] args) {
		System.out.println(5/2);
		Scanner in=new Scanner(System.in);
		int times=in.nextInt();
		long min,max=0;
		for (int i=1;i<=times;i++) {
			min=in.nextInt();
			max=in.nextInt();
			if(min==max) {
				System.out.println(0);
			}else if ((min<=max/2)&&max%2==1) {
				System.out.println(max/2);
			}else if (min<=max/2&&max%2==0) {
				System.out.println(max/2-1);
			}else{
				System.out.println(max%min);
			}
		}
	}
}
