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.*;

public class Demo{
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();        
        for(int i=0; i<t; i++){
            long l = sc.nextInt();
            long r = sc.nextInt();
            // Long[][] dp = new Long[Integer.MAX_VALUE][Integer.MAX_VALUE];
            System.out.println(solve(l, r));
        }
    }
    public static long solve(long l, long r){
        
        if(l>= r/2){
            if(l==r/2) return r/2-1;
            else return r%l;
        }else{
            if(r%2 != 0){
                r++;
            }
           long x = r/2 + 1;
            return r % x;
        }

        // if(dp[(int) l][(int) r] != null) return dp[(int) l][(int) r];
        // // System.out.println(l +"  "+ r + " mode "+ mod);
        // return dp[(int) l][(int) r] =  Math.max(mod,
        //              Math.max(
        //                 solve(l+1, r, r%l, dp),
        //                 solve(l, r-1, r%l, dp)
        //             )
        //         );
    }        
}


