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;
const long long INF = 1e18;
long long modu(long long n) {
  if (n < 0)
    return -n;
  else
    return n;
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int32_t main() {
  fastio();
  long long t;
  cin >> t;
  while (t--) {
    long long n, x, m, left, right, l, r, count = 0, i;
    cin >> n >> x >> m;
    left = -1, right = -1;
    while (m--) {
      cin >> l >> r;
      if (left == -1 && right == -1) {
        if (x >= l && x <= r) {
          count += r - l + 1;
          left = l;
          right = r;
        }
      }
      if (l <= left && r >= right) {
        count += left - l;
        count += r - right;
        left = l;
        right = r;
      } else if (l >= left && l <= right) {
        if (r > right) {
          count += r - right;
          right = r;
        }
      } else if (r >= left && r <= right) {
        if (l < left) {
          count += modu(l - left);
          left = l;
        }
      }
    }
    if (count > 0)
      cout << count << "\n";
    else
      cout << 1 << "\n";
  }
}
