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.
T = int(input())

def findK(arr):
  for i in range(0,len(arr)):
        if contains1(arr[i],x):
            K = i
            return K
            
  return -1

def solve(arr,x,K):
        
  
    a,b = arr[K]
    s = arr[K]
    for i in range(K,len(arr)-1):

        
            

        if hasOverlap(s,arr[i+1]):
            
            a1, b1 = arr[i+1]
            ##print('asd', a,b,a1,b1)
            if a1<a: 

                a = a1
            if b1>b:
                b = b1

            s = [a,b]
            
    return b - a + 1
            
def contains1(i,x):

    c,d = i
    if c<=x<=d:
        return 1
    else:
        return 0

def hasOverlap(i1,i2):

    c1,d1 = i1
    c2,d2 = i2
    
    if c2<=c1<=d2: 
        return c2, max(d1,d2)
    if c1<=c2<=d1:
        return c1, max(d1,d2)
    return 0
    
while T>0:
    n,x,m = [int(el) for el in input().rstrip().split(' ')] 
    arr = []
    for i in range(0,m):
        c,d = [int(el) for el in input().rstrip().split(' ')] 
        
        arr.append([c,d])
        
    K = findK(arr)
    if K!=-1:
        ans=solve(arr,x,K)        
    else:
        ans=1
    print(ans)
    T-=1


