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.*;
import java.lang.*;
import static java.lang.Math.*;

// Sachin_2961 submission //
public class Codeforces {

    public void solve() {
        long l = fs.nLong(), r = fs.nLong();
        if( l == r){
            out.println("0");
            return;
        }
        if(2*l > r ){
            long ans = r%l;
            out.println(ans);
            return;
        }
        long ans = r/2 - (r%2==0?1:0);
        out.println(ans);
    }
    static boolean multipleTestCase = true; static FastScanner fs; static PrintWriter out;
    public void run(){
        fs = new FastScanner();
        out = new PrintWriter(System.out);

        int tc = (multipleTestCase)?fs.nInt():1;
        while (tc-->0)solve();
        out.flush();
        out.close();
    }
    public static void main(String[]args){
        try{
            new Codeforces().run();
        }catch (Exception e){
            e.printStackTrace();
        }

    }
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String n() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }
        String Line()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
        int nInt() {return Integer.parseInt(n()); }
        long nLong() {return Long.parseLong(n());}
        double nDouble(){return Double.parseDouble(n());}
        int[]aI(int n){
            int[]ar = new int[n];
            for(int i=0;i<n;i++)
                ar[i] = nInt();
            return ar;
        }
    }
    public static void sort(int[] arr){
        ArrayList<Integer> ls = new ArrayList<Integer>();
        for(int x: arr)
            ls.add(x);
        Collections.sort(ls);
        for(int i=0; i < arr.length; i++)
            arr[i] = ls.get(i);
    }
}