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 T>
istream &operator>>(istream &is, vector<T> &arr);
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr);
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p);
template <typename T>
istream &operator>>(istream &is, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a << " ";
  }
  os << endl;
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p) {
  os << p.first << " " << p.second << '\n';
  return os;
}
long long INF = 1e18 + 5;
long long MOD = 1e9 + 7;
long long mul(long long a, long long b, long long md = MOD) {
  return (a % md * b % md) % md;
}
long long add(long long a, long long b, long long md = MOD) {
  return (a % md + b % md) % md;
}
bool valid(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
vector<pair<int, int>> dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
long long binPow(long long x, long long y) {
  int ans = 1;
  while (y > 0) {
    if (y % 2 == 1) ans = mul(ans, x);
    x = mul(x, x);
    y /= 2;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, x, m;
    cin >> n >> x >> m;
    int l = x, r = x;
    for (int i = 0; i < int(m); i++) {
      int left, right;
      cin >> left >> right;
      if (right < l || left > r)
        continue;
      else {
        l = min(l, left);
        r = max(r, right);
      }
    }
    cout << r - l + 1 << endl;
  }
}
