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.
def pos_ind(n,x,m,ops):
	var=0
	for i in range(m):
		l=ops[i]
		if x<=l[1] and x>=l[0]:
			var=1
			ind=i
			break
	if var==0:
		return 1
	else:
		a=ops[ind][0]
		b=ops[ind][1]
		for i in range(ind,m-1):
			l=ops[i+1]
			if l[0]<=b and l[1]>=a:
				a=min(a,l[0])
				b=max(b,l[1])
		return (b-a+1)

def new_fun(n,x,m,ops):
	L=[0,]*n
	L[x-1]=1
	for l in ops:
		a=l[0]-1
		b=l[1]-1
		var=0
		for i in range(a,b+1):
			if L[i]==1:
				var=1
				break
		if var==1:
			for i in range(a,b+1):
				L[i]=1
	return L.count(1)
t=int(input())
for i in range(t):
	n,x,m=map(int,input().split())
	ops=[]
	for j in range(m):
		l=list(map(int,input().split()))
		ops.append(l)
	print(pos_ind(n,x,m,ops))