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;
void printArray(int a[], int n) {
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void printVector(vector<int> v) {
  for (auto x : v) cout << x << " ";
  cout << "\n";
}
string reversed(string s) {
  string t = "";
  for (int i = s.length() - 1; i >= 0; i--) t += s[i];
  return t;
}
string binary(long long int n, int val) {
  string s = "";
  while (n > 0) {
    if (n % 2 == 1)
      s += '1';
    else
      s += '0';
    n /= 2;
  }
  int k = val - s.size();
  for (int i = 0; i < k; i++) s += '0';
  return reversed(s);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  cin >> t;
  for (int i1 = 0; i1 < t; i1++) {
    long long int n, x, m, l, r;
    cin >> n >> x >> m;
    l = x;
    r = x;
    for (int i = 0; i < m; i++) {
      long long int a, b;
      cin >> a >> b;
      if ((l >= a && l <= b) || (r >= a && r <= b)) {
        l = min(l, a);
        r = max(r, b);
      }
    }
    cout << r - l + 1 << "\n";
  }
}
