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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class a {
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        int tc = Integer.parseInt(br.readLine());

        for (int tt = 0; tt < tc; tt++) {
            st = new StringTokenizer(br.readLine());
            int l = Integer.parseInt(st.nextToken());
            int r = Integer.parseInt(st.nextToken());

            int answer = 0;
            for (int div = (l + r) / 2 - 10; div <= (l + r) / 2 + 10; div++) {
                if (div > r || div < l) continue;
                answer = Math.max(answer, r % div);
            }

            for (int div = r / 2 - 10; div <= r / 2 + 10; div++) {
                if (div > r || div < l) continue;
                answer = Math.max(answer, r % div);
            }

            System.out.println(answer);
        }
    }


    private static int gcd(int a, int b) {
        if (a % b == 0) return b;
        else return gcd(b, a % b);
    }
}