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.
#include <bits/stdc++.h>
using namespace std;
int char_to_int(char c) {
  int x = c - 48;
  return x;
}
char int_to_char(int x) {
  int c = x + 48;
  return c;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
map<int, int> mp;
vector<int> v1, v2;
map<int, vector<int> > G;
map<int, int> visited;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
int main() {
  fast();
  int tst;
  cin >> tst;
  while (tst--) {
    long long n, x, m, a, b;
    cin >> n >> x >> m;
    long long ans = 0LL;
    long long p = x, q = x;
    bool flag = false;
    for (int i = 1; i <= m; i++) {
      long long a, b;
      cin >> a >> b;
      if (!flag) {
        if (x >= a && x <= b) flag = true, p = a, q = b;
        continue;
      }
      if (flag && (p >= a && p <= b) || (q >= a && q <= b)) {
        if (a < p) p = a;
        if (b > q) q = b;
      }
    }
    cout << (q - p) + 1 << endl;
  }
  return 0;
}
