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.
#!/usr/bin/env python3

def intersects1(l, r, interval):
	(x, y) = interval
	low = max(l, x)
	high = min(r, y)

	return low <= high

def intersects2(l, r, intervals):
	for interval in intervals:
		if intersects1(l, r, interval):
			return True
	return False

def count(intervals):
	changes = []
	for (a, b) in intervals:
		changes += [(a, 1)]
		changes += [(b+1, -1)]

	changes = sorted(changes)

	num_intervals = 0
	last_pos = None

	ret = 0

	for (pos, delta) in changes:
		if num_intervals > 0:
			if last_pos == pos:
				continue	
			# print(f"counting [{last_pos}, {pos})")
			ret += (pos - last_pos)
			
		num_intervals += 1
		last_pos = pos

	return ret


def ans(n, x, lr):
	
	possible = [(x, x)]

	for (l, r) in lr:
		if intersects2(l, r, possible):
			possible += [(l, r)]

	return count(possible)

T = int(input())
for t in range(T):
	n, x, m = input().split(' ')
	n = int(n)
	x = int(x)
	m = int(m)
	lr = []
	for _ in range(m):
		l, r = input().split(' ')
		l = int(l)
		r = int(r)
		lr += [(l, r)]
	print(ans(n, x, lr))