#P1366B. Shuffle

Shuffle

No submission language available for this problem.

Description

You are given an array consisting of nn integers a1a_1, a2a_2, ..., ana_n. Initially ax=1a_x = 1, all other elements are equal to 00.

You have to perform mm operations. During the ii-th operation, you choose two indices cc and dd such that lic,dril_i \le c, d \le r_i, and swap aca_c and ada_d.

Calculate the number of indices kk such that it is possible to choose the operations so that ak=1a_k = 1 in the end.

The first line contains a single integer tt (1t1001 \le t \le 100) — the number of test cases. Then the description of tt testcases follow.

The first line of each test case contains three integers nn, xx and mm (1n1091 \le n \le 10^9; 1m1001 \le m \le 100; 1xn1 \le x \le n).

Each of next mm lines contains the descriptions of the operations; the ii-th line contains two integers lil_i and rir_i (1lirin1 \le l_i \le r_i \le n).

For each test case print one integer — the number of indices kk such that it is possible to choose the operations so that ak=1a_k = 1 in the end.

Input

The first line contains a single integer tt (1t1001 \le t \le 100) — the number of test cases. Then the description of tt testcases follow.

The first line of each test case contains three integers nn, xx and mm (1n1091 \le n \le 10^9; 1m1001 \le m \le 100; 1xn1 \le x \le n).

Each of next mm lines contains the descriptions of the operations; the ii-th line contains two integers lil_i and rir_i (1lirin1 \le l_i \le r_i \le n).

Output

For each test case print one integer — the number of indices kk such that it is possible to choose the operations so that ak=1a_k = 1 in the end.

Samples

Sample Input 1

3
6 4 3
1 6
2 3
5 5
4 1 2
2 4
1 2
3 3 2
2 3
1 2

Sample Output 1

6
2
3

Note

In the first test case, it is possible to achieve ak=1a_k = 1 for every kk. To do so, you may use the following operations:

  1. swap aka_k and a4a_4;
  2. swap a2a_2 and a2a_2;
  3. swap a5a_5 and a5a_5.

In the second test case, only k=1k = 1 and k=2k = 2 are possible answers. To achieve a1=1a_1 = 1, you have to swap a1a_1 and a1a_1 during the second operation. To achieve a2=1a_2 = 1, you have to swap a1a_1 and a2a_2 during the second operation.