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.
# May the Speedforce be with us...
'''
problem statement achhe se padhna hai
age se bhi dekhna hai, piche se bhi dekhna hai
'''
#from math import gcd,ceil,sin,radians,cos
#from bisect import insort
#from collections import defaultdict as dd
def lcm(a,b):
	return (a*b)//gcd(a,b)
def inin():
	return int(input())
def inar():
	return list(map(int,input().split()))
def ar(element,size):
	return [element for i in range(size)]
def yes():
	print('Yes')
def no():
	print('No')
'''
def countDivisibles(A, B, M): #count numbers in the range[A,B] which are divisible by M
    if (A % M == 0): 
        return ((B // M) - (A // M)) + 1
    return ((B // M) - (A // M)) 
	
def  countSetBits(n): #count the number of set bits in the binary representation of an integer
    count = 0
    while (n): 
        count += n & 1
        n >>= 1
    return count
def digitsum(num): #return the sum of dogots of a number
	su=0
	while(num):
		su+=num%10
		num//=10
	return su
'''


Testcase=inin()
for _ in range(Testcase):
	n,x,m=inar()
	p,q=x,x
	for i in range(m):
		l,r=inar()
		if l<p and q<r:
			p=l 
			q=r 
			continue
		if p<=l<=q and r>q:
			q=r 
			continue
		if l<p and p<=r<=q:
			p=l
	print(q-p+1)


	
