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 A {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

    int t = Integer.parseInt(br.readLine());
    for (int i = 0; i < t; i++) {
      String[] inputs = br.readLine().split(" ");
      int l = Integer.parseInt(inputs[0]);
      int r = Integer.parseInt(inputs[1]);
      // r >= a >= b >= l

      int mod2 = (r+2) % 2;
      int newR = r + mod2;
      int halfR = newR/2;
      if (l <= halfR) {
        pw.println(halfR - 1);
      } else {
        pw.println(r % l);
      }
    }

    pw.flush();
  }
}
