You are given an array consisting of nn integers a1a1, a2a2, ..., anan. Initially ax=1ax=1, all other elements are equal to 00.

You have to perform mm operations. During the ii-th operation, you choose two indices cc and dd such that li≤c,d≤rili≤c,d≤ri, and swap acac and adad.

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

Input

The first line contains a single integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then the description of tt testcases follow.

The first line of each test case contains three integers nn, xx and mm (1≤n≤1091≤n≤109; 1≤m≤1001≤m≤100; 1≤x≤n1≤x≤n).

Each of next mm lines contains the descriptions of the operations; the ii-th line contains two integers lili and riri (1≤li≤ri≤n1≤li≤ri≤n).

Output

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

题目类型：没啥啊类型

解题目标：输出有多少个可能为1的点。

解题思路：不断扩充左边界与右边界，最后输出r-l+1.

AC代码：

#include  <bits/stdc++.h>
using namespace std;

int main()
{
	int t;
	scanf("%d", &t);
	while(t -- )
	{
		int n, m, x;
		scanf("%d%d%d", &n, &x, &m);
		int l = x, r=  x;
		int temp1, temp2;
		rep(i, 1, m)
		{
			scanf("%d%d", &temp1, &temp2);
			if( l>= temp1 && l<= temp2) l = temp1;
			if( r>= temp1 && r<= temp2) r =temp2;
		}
		 cout<<r-l+1<<endl;
	 } 
	return 0;
}