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.*;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        int t = nextInt();
        for (int i = 0; i < t; i++) {
            int l = nextInt();
            int r = nextInt();
            if (l <= r / 2 + 1) out.println(r % (r / 2 + 1));
            else out.println(r % l);
        }
        out.close();
    }

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter out = new PrintWriter(System.out);
    static StringTokenizer in = new StringTokenizer("");


    public static boolean hasNext() throws IOException {
        if (in.hasMoreTokens()) return true;
        String s;
        while ((s = br.readLine()) != null) {
            in = new StringTokenizer(s);
            if (in.hasMoreTokens()) return true;
        }
        return false;
    }

    public static String nextToken() throws IOException {
        while (!in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public static int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }
}