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.*;
import java.io.*;

public class A {
    public static void main(String[] args) throws IOException {
        MyScanner sc = new MyScanner();
        int T = sc.nextInt();
        while(T-- > 0) {
            int l = sc.nextInt();
            int r = sc.nextInt();
            l = Math.max(r / 2, l) + 1;
            System.out.println(r % l);
        }
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer tok;
    MyScanner() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        tok = new StringTokenizer("");
    }
    public String next() throws IOException {
        while(!tok.hasMoreTokens()) tok = new StringTokenizer(br.readLine());
        return tok.nextToken();
    }
    public int nextInt() throws IOException {
        return Integer.parseInt(next());
    }
}

