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.BufferedReader;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class Main {
    public static void main(String args[])throws Exception{
        Input sc=new Input();
        StringBuilder sb=new StringBuilder();
        int t=sc.readInt();
        for(int f=0;f<t;f++){
            int d[]=sc.readArray();
            int l=d[0];
            int r=d[1];
            int n=r/2;
            if(l>n){
               sb.append(r%l);
            }else{
                sb.append(r%(n+1));
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
}
class Input{
    BufferedReader br;
    StringTokenizer st;
    Input(){
        br=new BufferedReader(new InputStreamReader(System.in));
        st=new StringTokenizer("");
    }
    public int[] readArray() throws Exception{
        st=new StringTokenizer(br.readLine());
        int a[]=new int[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Integer.parseInt(st.nextToken());
        }
        return a;
    }
    public long[] readArrayLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        long a[]=new long[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Long.parseLong(st.nextToken());
        }
        return a;
    }
    public int readInt() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }
    public long readLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Long.parseLong(st.nextToken());
    }
    public String readString() throws Exception{
        return br.readLine();
    }
    public int[][] read2dArray(int n,int m)throws Exception{
        int a[][]=new int[n][m];
        for(int i=0;i<n;i++){
            st=new StringTokenizer(br.readLine());
            for(int j=0;j<m;j++){
                a[i][j]=Integer.parseInt(st.nextToken());
            }
        }
        return a;
    }
}


