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 int N = 2e5 + 1;
long long int power(long long int x, long long int y, long long int p) {
  long long int temp;
  if (y <= 0) return 1;
  temp = power(x, y / 2, p);
  if (y % 2 == 0) {
    return (temp * temp) % p;
  } else {
    return (((x * temp) % p) * temp) % p;
  }
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (r == 0) {
    return 1;
  }
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % p;
  }
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
void leftRotatebyOne(long long int arr[], int n) {
  int temp = arr[0], i;
  for (i = 0; i < n - 1; i++) arr[i] = arr[i + 1];
  arr[i] = temp;
}
void leftRotate(long long int arr[], int d, int n) {
  for (int i = 0; i < d; i++) leftRotatebyOne(arr, n);
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  long long int t;
  cin >> t;
  for (long long int j = 0; j < t; j++) {
    long long int n, x, m;
    cin >> n >> x >> m;
    long long int left = x;
    long long int right = x;
    while (m--) {
      long long int l, r;
      cin >> l >> r;
      if (!(l > right || r < left)) {
        left = min(left, l);
        right = max(right, r);
      }
    }
    cout << right - left + 1 << endl;
  }
  return 0;
}
