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;
template <typename a>
void SWAP(a &b, a &c) {
  a d = b;
  b = c;
  c = d;
}
long long power(long long a, long long b) {
  long long result = 1;
  while (b) {
    if (b & 1) result = a * result;
    a *= a;
    b >>= 1;
  }
  return result;
}
long long gcd(long long a, long long b) {
  while (b) {
    a = a % b;
    SWAP(a, b);
  }
  return a;
}
long long lcm(long long a, long long b) { return (a * (b / gcd(a, b))); }
long long check_double_equality(double a, double b) {
  double c = (((a - b) > 0) ? (a - b) : (-(a - b)));
  if (c < 1e-9)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, x, m;
    cin >> n >> x >> m;
    long long l, r;
    long long left = 1, right = 1;
    bool one_found = false;
    for (long long i = 0; i < m; i++) {
      cin >> l >> r;
      if (l <= x && x <= r && !one_found) {
        left = l;
        right = r;
        one_found = true;
      }
      if (one_found) {
        if ((left <= l && l <= right) || (left <= r && r <= right) ||
            (l <= left && left <= r) || (l <= right && right <= r)) {
          left = (((left) < (l)) ? (left) : (l));
          right = (((right) > (r)) ? (right) : (r));
        }
      }
    }
    long long ans = right - left + 1;
    cout << (ans) << '\n';
  }
  return 0;
}
