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.
//package codeforces;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class cf
{
	public static void main(String[]args) {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		while(t-->0) {
			
			int l = sc.nextInt();
			int r = sc.nextInt();
			if(l==r) {System.out.println(0);continue;}
			int mid = r/2;
			mid++;
			
			if(l<=mid && mid<=r)
				System.out.println(r%mid);
			else {
				int res = Math.max(r%l,r%(r-1));
				System.out.println(res);
			}
			
			
		}
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int [] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		long [] longArray(int n) {
			long[] a=new long[n];
			for(int i=0 ; i<n ; i++) a[i]=nextLong();
			return a;
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}


}

