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.
//package test.codeforce.edu89;

import java.util.Scanner;

public class B
{

	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int testNum = input.nextInt();
		StringBuffer stringBuffer = new StringBuffer();
		for (int rr = 0; rr < testNum; rr++)
		{
			int n = input.nextInt();
			int x = input.nextInt();
			int m = input.nextInt();
			int[][] an = new int[m][2];
			for (int i = 0; i < m; i++)
			{
				an[i][0] = input.nextInt();
				an[i][1] = input.nextInt();
			}
			int ans = 1;
			int left=x;
			int right=x;
			for (int i = 0; i < m; i++)
			{
				if (an[i][0] <=right && an[i][1] >= left)
				{
					if (right >= an[i][1] && left > an[i][0])
					{
						ans+=left-an[i][0];
					}
					if (right <an[i][1] && left <=an[i][0])
					{
						ans+=an[i][1]-right;
					}
					if(right <an[i][1]&&left > an[i][0]){
						ans+=an[i][1]-right+left-an[i][0];
					}
					left = Math.min(left,an[i][0]);
					right = Math.max(right, an[i][1] );
				}
			}
			//������������

			stringBuffer.append(ans + "\n");
		}
		System.out.println(stringBuffer.toString());
	}
}
