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.*;

/* Name of the class has to be "Main" only if the class is public. */
public class q2
{
	public static void main (String[] args) throws java.lang.Exception
	{
        // your code goes here
        Scanner sc = new Scanner(System.in);
        int t= Integer.parseInt(sc.next());
        while(t-->0)
        {
            long n = Long.parseLong(sc.next());
            long x = Long.parseLong(sc.next());
            long m = Long.parseLong(sc.next());
            long c1=x;long c2=x; 
            long count=0;
            int flag=0;
            long ans1=0;
            long ans2=0;
            for(long i=0;i<m;i++)
            {
                long a = Long.parseLong(sc.next());
                long b =Long.parseLong(sc.next());

                if(x<=b && x>=a && flag==0)
                {
                    c1=a;
                    c2=b;
                    flag=1;
                    count=(b-a+1);
                }
                if((( b>=c2 && a<=c2)||(a<=c1 && b>=c1))&& flag==1)
                {
                    //ans1= Math.max(b,c2)-Math.min(a,c1)+1;
                    ans2= Math.max(b,c2)-Math.min(a,c1)+1;
                    c1=Math.min(a,c1);
                    c2= Math.max(b,c2);
                    //System.out.println(c1+" "+c2);
                }

            }
            System.out.println(c2-c1+1);
           // System.out.println(count>ans2?count:ans2);
        }
    }
}