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.* ;
import java.util.* ;


 
public class App {
    public static void main (String[] args) throws java.lang.Exception {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        //reader = new BufferedReader(new FileReader(new File("C:\\Users\\Anjali\\Desktop\\projects\\HelloWorld\\src\\input.txt")));
		BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
       // writer = new BufferedWriter(new FileWriter(new File("C:\\Users\\Anjali\\Desktop\\projects\\HelloWorld\\src\\output.txt")));
		//int tests = Integer.parseInt(reader.readLine());
        int tests = 1;
		//StringBuilder sb = new StringBuilder(tests);
		while (tests-- > 0){
		    solve(reader, writer);
            writer.write("\n");
		}
		reader.close();
        writer.flush();
        writer.close();
    }

 
    static void solve(BufferedReader reader, BufferedWriter writer) throws  IOException {
		int tests = Integer.parseInt(reader.readLine());
		for (int i = 0; i < tests; i++) {
			String[] input = reader.readLine().split(" ");
			long l = Long.parseLong(input[0]);
			long r = Long.parseLong(input[1]);
			long ans = 0;
			if (l <= (r / 2) + 1) {
				ans = (r - 1) / 2;
			} else {
				ans = r - l;
			}
			writer.write(ans + "\n");
		}	
	}
}
