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.
import os
### START FAST IO ###
os_input = os.read(0, int(1e7)).split()
os_input_pos = -1
answer_list = []
def read_s():
    global os_input_pos
    os_input_pos += 1
    return os_input[os_input_pos].decode()
def read_i():
    return int(read_s())
def write_s(v):
    answer_list.append(v)
def write_i(v):
    write_s(str(v))
def print_ans():
    os.write(1, "\n".join(answer_list).encode())
    os.write(1, "\n".encode())
#### END FAST IO ####

T = read_i()
while T:
  T -= 1
  n, x, m = [read_i() for i in range(3)]
  left = x
  right = x
  for i in range(m):
    l = read_i()
    r = read_i()
    if left <= r and l <= right:
      left = min(left, l)
      right = max(right, r)
  write_i(right - left + 1)

print_ans()

