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>
unsigned const int m = 1e9 + 7;
using namespace std;
inline long long axlb(long long a, long long b) {
  return (b % a == 0 ? (b / a) - 1 : b / a);
}
inline long long axgeb(long long a, long long b) {
  return (b % a == 0 ? b / a : (b / a) + 1);
}
inline long long axgb(long long a, long long b) { return (b / a) + 1; }
inline long long axleb(long long a, long long b) { return b / a; }
inline void printyn(bool c) {
  cout << (c ? "YES" : "NO") << "\n";
  return;
}
template <typename T>
inline void scanarr(T* arr, long long n) {
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
  }
}
template <typename T>
void scanvec(vector<T>* vec, long long n) {
  vec->reserve(n + 10);
  for (long long i = 0; i < n; i++) {
    T temp;
    cin >> temp;
    vec->push_back(temp);
  }
  return;
}
template <typename T>
inline void printarr(T* arr, long long n) {
  for (long long i = 0; i < n; i++) {
    cout << arr[i] << " ";
  }
  cout << "\n";
}
template <typename T>
inline void printvec(vector<T>* vec) {
  long long n = vec->size();
  for (long long i = 0; i < n; i++) {
    cout << vec->at(i) << " ";
  }
  cout << "\n";
}
void solve() {}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T;
  cin >> T;
  while (T--) {
    long long n, x, m;
    cin >> n >> x >> m;
    long long l = -1, r = -1;
    while (m--) {
      long long li, ri;
      cin >> li >> ri;
      if (l == -1 && r == -1 && x >= li && x <= ri) {
        l = li;
        r = ri;
      }
      if (li < l && ri < l) {
        continue;
      }
      if (li > r && ri > r) {
        continue;
      } else {
        l = (l > li ? li : l);
        r = (ri > r ? ri : r);
      }
    }
    long long ans = (l == -1 && r == -1) ? 1 : r - l + 1;
    cout << ans << "\n";
  }
}
