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;
bool isPrime(int x) {
  if (x == 1) return false;
  for (long long i = 2; i <= sqrt(x); i += 1)
    if (x % i == 0) return false;
  return true;
}
void print(int a[], int n) {
  for (long long i = 0; i < n; i += 1) cout << a[i] << " ";
}
void iarr(int *a, int n) {
  for (long long i = 0; i < n; i += 1) cin >> a[i];
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 1000000007) * (a % 1000000007)) % 1000000007;
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    b >>= 1;
  }
  return res;
}
long long numFactors(long long x) {
  long long cnt = 0;
  for (long long i = 1; i <= sqrt(x); i += 1) {
    if (x % i == 0) {
      if (x / i == i)
        cnt += 1;
      else
        cnt += 2;
    }
  }
  return cnt;
}
int main() {
  long long T;
  cin >> T;
  while (T--) {
    int n, x, m;
    cin >> n >> x >> m;
    std::vector<pair<int, int> > v;
    int s = INT_MAX, e = INT_MIN;
    int t = 0, f = 0, ind = 0;
    while (t < m) {
      int a, b;
      cin >> a >> b;
      v.push_back(make_pair(a, b));
      if (a <= x and x <= b and f == 0) {
        s = min(s, a);
        e = max(e, b);
        f = 1;
        ind = t;
      }
      t++;
    }
    if (f == 0) {
      cout << 1 << "\n";
      continue;
    }
    for (long long i = ind + 1; i < m; i += 1) {
      if ((v[i].first <= s and v[i].second >= s) or
          (v[i].first <= e and v[i].second >= e)) {
        s = min(s, v[i].first);
        e = max(e, v[i].second);
      }
    }
    cout << e - s + 1 << "\n";
  }
  return 0;
}
