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.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;

import static java.lang.Math.max;
import static java.lang.Math.min;

public class Main {

    public static void main(String[] args) {
        try(PrintWriter pw = new PrintWriter(System.out)) {
            try (Scanner sc = new Scanner(System.in)) {
                solve(pw, sc);
            }
        }
    }

    private static void solve(PrintWriter out, Scanner in) {
        preCalc();
        int t = in.nextInt();
        in.nextLine();
        for (int i = 0; i < t; i++) {
            solveNext(in, out);
        }
    }

    private static void preCalc() {

    }

    private static void solveNext(Scanner in, PrintWriter out) {
        int l = in.nextInt();
        int r = in.nextInt();
        out.println(r % max(l, r / 2 + 1));
    }

}
