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 M = 1e9 + 9;
const int MaxM = 100005;
void solveA() {
  string s;
  cin >> s;
  s += 'c';
  int n = s.size();
  vector<int> v;
  int d = 0, e = 0, s1 = 0, s2 = 0, s3 = 0;
  for (int i = 0; i < n; i++) {
    if (s[i] == 'b') {
      d++;
      s1 = max(s1, d);
    } else {
      d = 0;
    }
    if (s[i] == 'a') {
      e++;
      s2 = max(s2, e);
    } else {
      v.push_back(e);
      e = 0;
    }
  }
  sort(v.begin(), v.end());
  if (v.size() >= 2) {
    s3 = v[v.size() - 2];
  }
  cout << s1 + s2 + s3;
}
void solve() {
  int n;
  cin >> n;
  int m, x;
  cin >> x >> m;
  int ans, a = x, b = x;
  vector<pair<int, int>> v(m);
  for (int i = 0; i < m; i++) {
    cin >> v[i].first;
    cin >> v[i].second;
  }
  for (int i = 0; i < m; i++) {
    if (v[i].first <= a && v[i].second >= a) {
      a = v[i].first;
    }
    if (v[i].second >= b && v[i].first <= b) {
      b = v[i].second;
    }
  }
  ans = b - a + 1;
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  vector<pair<int, int>> v;
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}
