You are given an array consisting of n integers a_1, a_2, ..., a_n. Initially a_x = 1, all other elements are equal to 0.

You have to perform m operations. During the i-th operation, you choose two indices c and d such that l_i ≤ c, d ≤ r_i, and swap a_c and a_d.

Calculate the number of indices k such that it is possible to choose the operations so that a_k = 1 in the end.

Input

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Then the description of t testcases follow.

The first line of each test case contains three integers n, x and m (1 ≤ n ≤ 10^9; 1 ≤ m ≤ 100; 1 ≤ x ≤ n).

Each of next m lines contains the descriptions of the operations; the i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).

Output

For each test case print one integer — the number of indices k such that it is possible to choose the operations so that a_k = 1 in the end.

Example

Input


3
6 4 3
1 6
2 3
5 5
4 1 2
2 4
1 2
3 3 2
2 3
1 2


Output


6
2
3

Note

In the first test case, it is possible to achieve a_k = 1 for every k. To do so, you may use the following operations:

  1. swap a_k and a_4; 
  2. swap a_2 and a_2; 
  3. swap a_5 and a_5. 



In the second test case, only k = 1 and k = 2 are possible answers. To achieve a_1 = 1, you have to swap a_1 and a_1 during the second operation. To achieve a_2 = 1, you have to swap a_1 and a_2 during the second operation.
import java.util.*;
import java.lang.*;
import java.io.*;
 

public class B
{
	public static void main (final String[] args) throws java.lang.Exception {
        final Scanner sc = new Scanner(System.in);
        final int t = sc.nextInt();
        for(int i=0; i<t; i++){
            solution(sc);
        }
    }

    private static void solution(Scanner sc){
        int n = sc.nextInt();
        int x = sc.nextInt();
        int m = sc.nextInt();
        int lo = x;
        int hi = x;
        for(int i=0; i<m; i++){
            int iv_lo = sc.nextInt();
            int iv_hi = sc.nextInt();
            if((hi >= iv_lo && lo <= iv_hi) || (lo <= iv_hi && hi >= iv_lo)){
                lo = Math.min(lo, iv_lo);
                hi = Math.max(hi, iv_hi);
            }
        }
        System.out.println(hi-lo+1);
    }

}