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>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
long long int ll_max(long long int a, long long int b, long long int c) {
  return max(a, max(b, c));
}
const int inf = 0x3f3f3f3f;
const long long int llinf = 0x3f3f3f3f3f3f;
const int mod = 1000000007;
const char nl = '\n';
const int MX = 100001;
int64_t solve() {
  int64_t n, x, m;
  cin >> n >> x >> m;
  int64_t maxm = 0, l, r;
  cin >> l >> r;
  int64_t rlo = 0, rhi = 0;
  if (l <= x && x <= r) {
    maxm = r - l + 1;
    rlo = l;
    rhi = r;
  }
  m--;
  while (m--) {
    cin >> l >> r;
    if (maxm == 0) {
      if (l <= x && x <= r) {
        maxm = r - l + 1;
        rlo = l;
        rhi = r;
      }
    } else {
      if (l > rhi || r < rlo) continue;
      int64_t lo, hi;
      (l < rlo) ? lo = l : lo = rlo;
      (r > rhi) ? hi = r : hi = rhi;
      maxm = max(maxm, hi - lo + 1);
      rlo = lo;
      rhi = hi;
    }
  }
  if (maxm == 0) maxm = 1;
  return maxm;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int tt;
  cin >> tt;
  while (tt--) {
    cout << solve() << endl;
  }
  return 0;
}
