B. Shuffle

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Example

input

Copy

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

Copy

6
2
3

Note

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

1. swap akak and a4a4;
2. swap a2a2 and a2a2;
3. swap a5a5 and a5a5.

In the second test case, only k=1k=1 and k=2k=2 are possible answers. To achieve a1=1a1=1, you have to swap a1a1 and a1a1 during the second operation. To achieve a2=1a2=1, you have to swap a1a1 and a2a2 during the second operation.

解題說明：水題，只需要找到能交換數字的區間，然後計算即可。

#include<stdio.h>

int main()
{
	int t, m;
	long int n, x, l, r, ma, mi;
	scanf("%d", &t);
	while (t--) 
	{
		scanf("%ld %ld %d", &n, &x, &m);
		ma = x;
		mi = x;
		for (int i = 0; i<m; i++)
		{
			scanf("%ld %ld", &l, &r);
			if (mi <= r)
			{
				if (mi > l)
				{
					mi = l;
				}
			}
			if (ma >= l) 
			{
				if (ma < r)
				{
					ma = r;
				}
			}
		}
		printf("%ld\n", ma - mi + 1);
	}
	return 0;
}