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 N = pow(2, 32);
map<long long, long long> factorize(long long n) {
  map<long long, long long> ans;
  for (long long i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      ans[i]++;
      n /= i;
    }
  }
  if (n > 1) {
    ans[n]++;
    n = 1;
  }
  return ans;
}
void FAST() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long answer(long long sum, vector<vector<long long>> a, long long n,
                 long long m) {
  long long cnt1 = 0, cnt0 = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      if (i + j == sum || i + j == n + m - 2 - sum) {
        if (a[i][j])
          cnt1++;
        else
          cnt0++;
      }
    }
  }
  return min(cnt0, cnt1);
}
int main() {
  FAST();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long x;
    cin >> x;
    long long m;
    cin >> m;
    long long l = x, r = x;
    long long tl, tr;
    bool act = 0;
    while (m--) {
      cin >> tl >> tr;
      if (!(tr < l || tl > r)) {
        l = min(l, tl);
        r = max(r, tr);
      }
    }
    cout << r - l + 1 << "\n";
  }
}
