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 codeforces{
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static int[] INT(int N)throws IOException{
        int[] A = new int[N];
        String[] st = br.readLine().split(" ");
        for(int i=0;i<N;i++)
        A[i]  = Integer.parseInt(st[i]);
        return A;
    }

    public static void main(String[] args) throws IOException{
        StringBuilder res = new StringBuilder();
        int t = Integer.parseInt(br.readLine());
        while(t-->0){
            int[] arr = INT(2);
            int l = arr[0];
            int r = arr[1];
            int max = (r-1)/2;
            int allowed = 0;
            if(r%2==0)allowed = r/2 - 1;
            else allowed = r/2 + 1;
            if(l<=allowed)res.append(max+"\n");
            else res.append((r-l)+"\n");
        }
        System.out.println(res);
    }
}
            
            