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 Main {
    // For fast input output
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            try {
                br = new BufferedReader(new FileReader("input.txt"));
                PrintStream out = new PrintStream(new FileOutputStream("output.txt"));
                System.setOut(out);
            } catch (Exception e) {
                br = new BufferedReader(new InputStreamReader(System.in));
            }
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        int [] readintarray(int n) {
            int res [] = new int [n];
            for(int i = 0; i<n; i++)res[i] = nextInt();
            return res;
        }
        long [] readlongarray(int n) {
            long res [] = new long [n];
            for(int i = 0; i<n; i++)res[i] = nextLong();
            return res;
        }
    }
    static  int gcd(int a,int b)
    {
        if(b==0)
        {
            return a;
        }
        return gcd(b,a%b);
    }
 
    static int lcm(int a, int b)
    {
        return (a / gcd(a, b)) * b;
    }
 
 
    static int abs(int a)
    {
        if(a<0)
            return -1*a;
        return a;
    }
    public static int lower(int arr[],int key){
        int low = 0;
        int high = arr.length-1;
        while(low < high){
            int mid = low + (high - low)/2;
            if(arr[mid] >= key){
                high = mid;
            }
            else{
                low = mid+1;
            }
        }
        return low;
    }
    public static int upper(int arr[],int key){
        int low = 0;
        int high = arr.length-1;
        while(low < high){
            int mid = low + (high - low+1)/2;
            if(arr[mid] <= key){
                low = mid;
            }
            else{
                high = mid-1;
            }
        }
        return low;
    }

    // end of fast i/o code
    public static void main(String[] args) {
        FastReader reader = new FastReader();
         StringBuilder ans = new StringBuilder();   
        int t = reader.nextInt();
        while (t-- > 0) {
            int l = reader.nextInt();
            int r = reader.nextInt();

            if((r&1) == 1){
                if(r/2 < l){
                    ans.append(r%l + "\n");
                } else 
                ans.append(r/2 + "\n");
            
            } else {
                if(r/2 < l)
                 ans.append(r%l + "\n");
                 else 
                 ans.append(r/2-1 + "\n");
            }

            
        }
        System.out.println(ans);
    }
    
  
}