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.lang.*;
import java.io.*;
import java.util.Arrays;

public class Main
{
    public static void main (String[] args) throws java.lang.Exception
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int t;
        try {
            t = Integer.parseInt(br.readLine());
        } catch (Exception ignored){
            return;
        }
        while (t --> 0) {
            try {
                int[] arr = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
                if ((arr[1] & 1) == 1) {
                    if (arr[1] / 2 + 1 >= arr[0]) {
                        System.out.println(arr[1] / 2);
                    } else {
                        System.out.println(arr[1] % arr[0]);
                    }
                } else {
                    if (arr[1] / 2 + 1 >= arr[0]) {
                        System.out.println(arr[1] /2 - 1);
                    } else {
                        System.out.println(arr[1] % arr[0]);
                    }
                }
            } catch (Exception ignored){}
        }
    }
} 