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.
// Author: FarazAhmad
import java.util.*;
import java.io.*;
import java.math.*;
public class Fa {
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			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;
		}
	}
	static void swap(int []list,int i){
		int temp;
		temp=list[i];
		list[i]=list[i+1];
		list[i+1]=temp;
		return;
	}
static void decimal_places(double n){
	System.out.format("%.9f", n);
	System.out.println();
}
static void helper(char[] arr, int i, int prev){
	for(int j=i-1;j>=0;j--){
	  if(prev=='B'){
		arr[j]='R';
		prev='R';
	  }
	  else{
		arr[j]='B';
		prev='B';
	  }
	}
  }
  
	public static void main(String[] args)
	{
		FastReader sc = new FastReader();
		int t=sc.nextInt();
	// Vector<String> powers=new Vector<>();
	// BigInteger in =  new BigInteger("1000000000000000000");
	// 	   BigInteger ind =  new BigInteger("1");  
	// 	   BigInteger ii =  new BigInteger("2");  

	// 		while(ind.compareTo(in)==-1||ind.compareTo(in)==0)
	// 		{
	// 			powers.add(""+ind);
	// 			ind =ind.multiply(ii);
	// 		}
		//first:
        for(int tt=0;tt<t;tt++){ 
			long l=sc.nextLong();
			long r=sc.nextLong();
			if((r/2+1)>=l) System.out.println(r%(r/2+1));
			else System.out.println(r-l);
			

}
}
}

