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 mod = pow(10, 9) + 7;
const double pi = 3.141592653589793238;
long long countdivisors(long long n) {
  long long count = 0;
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i) {
        count++;
      } else {
        count = count + 2;
      }
    }
  }
  return count;
}
void remove(std::vector<int> &v) {
  auto end = v.end();
  for (auto it = v.begin(); it != end; ++it) {
    end = std::remove(it + 1, end, *it);
  }
  v.erase(end, v.end());
}
bool isPerfectSquare(long long x) {
  long long s = sqrt(x);
  return (s * s == x);
}
bool isFibonacci(int n) {
  return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);
}
int digSum(int n) {
  int sum = 0;
  while (n > 0) {
    sum += n % 10;
    n = n / 10;
  }
  return sum;
}
long long exponentMod(long long A, long long B, long long C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long)((y + C) % C);
}
void SieveOfEratosthenes(int n, bool isPrime[]) {
  for (int i = 0; i <= n; ++i) {
    isPrime[i] = true;
  }
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i * i <= n; ++i) {
    if (isPrime[i] == true) {
      for (int j = i * i; j <= n; j += i) isPrime[j] = false;
    }
  }
}
void solve() {
  long long n, x, m;
  cin >> n >> x >> m;
  long long l = x;
  long long r = x;
  while (m--) {
    long long a, b;
    cin >> a >> b;
    if ((a < l && b < l) || (a > r) && b > r) {
      continue;
    } else {
      l = min(l, a);
      r = max(r, b);
    }
  }
  cout << (r - l + 1) << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
