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 HelloWorld{

    private static PrintWriter pw = new PrintWriter(System.out);
    private static InputReader sc = new InputReader();
    private static final int intmax = Integer.MAX_VALUE, intmin = Integer.MIN_VALUE;
 
    static class InputReader{
        private static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        private static StringTokenizer tk;
 
        private void next()throws IOException{
            while(tk == null || !tk.hasMoreTokens())   
                tk = new StringTokenizer(r.readLine());
        }
 
        private int nextInt()throws IOException{
            next();
            return Integer.parseInt(tk.nextToken());
        }
 
        private long nextLong()throws IOException{
            next();
            return Long.parseLong(tk.nextToken());
        }
 
        private String readString()throws IOException{
            next();
            return tk.nextToken();
        }
 
        private double nextDouble()throws IOException{
            next();
            return Double.parseDouble(tk.nextToken());
        }
 
        private int[] intArray(int n)throws IOException{
            next();
            int arr[] = new int[n];
 
            for(int i=0; i<n; i++)
                arr[i] = nextInt();
 
            return arr;
        }
 
        private long[] longArray(int n)throws IOException{
            next();
            long arr[] = new long[n];
 
            for(int i=0; i<n; i++)
                arr[i] = nextLong();
 
            return arr;
        }
    }
 
    public static void main(String args[])throws IOException{
        int t = sc.nextInt();
        // int t = 1;
        while(t-->0)    solve();
        pw.flush();
        pw.close();
    }

    public static void solve()throws IOException{
        long l = sc.nextLong(), r = sc.nextLong();
        
        long begin = Math.max(r/2, l);
        
        if(l > r/2)
            pw.println(r - l);
        else
            pw.println((r + 1)/2 - 1);
        
        /*if(r == begin * 2)
            pw.println(begin - 1);
        else
            pw.println(r - begin);*/
    }
}