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.*;
public class Main{
    public static int maxRemainder(int l, int r){
    	int low=l;
    	int high=r;
    	int mid=0;
    	int maxAns=Integer.MIN_VALUE;
    	while(low<=high){
    		mid=(low+high)/2;
    		if(mid*2 <r){
    			maxAns=Math.max(maxAns,mid);
    			low=mid+1;
    		}
    		else{
    			high=mid-1;
    		}
    	}
    	if(maxAns>0) return maxAns;
    	else {
    		return r%l;
    	}
    }
    public static void main(String[] args){
        Scanner s=new Scanner(System.in);
        int t=s.nextInt();
        for(int i=0;i<t;i++){
        	int l=s.nextInt();
        	int r=s.nextInt();
        	System.out.println(maxRemainder(l,r));
        }
    }
}