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 int inf = 9e18;
const long double pi = 2 * acos(0.0);
long long int tes, n, x, m;
long long int power(long long int x, long long int y) {
  long long int res = 1ll;
  while (y > 0) {
    if (y & 1) res = res * x;
    y >>= 1;
    x = x * x;
  }
  return res;
}
void solve() {
  cin >> n >> x >> m;
  vector<pair<long long int, long long int>> arr(m);
  ;
  for (int i = 0; i < m; i++) cin >> arr[i].first >> arr[i].second;
  vector<bool> brr(m, false);
  ;
  for (int i = 0; i < m; i++) {
    if (x >= arr[i].first && x <= arr[i].second) brr[i] = true;
    ;
    for (int j = 0; j < i; j++) {
      if ((arr[i].first >= arr[j].first && arr[i].first <= arr[j].second) ||
          (arr[i].second >= arr[j].first && arr[i].second <= arr[j].second)) {
        if (brr[j]) brr[i] = true;
      }
      if (arr[j].first >= arr[i].first && arr[j].second <= arr[i].second) {
        if (brr[j]) brr[i] = true;
      }
    }
  }
  vector<pair<long long int, long long int>> crr;
  ;
  for (int i = 0; i < m; i++) {
    if (brr[i]) crr.push_back(arr[i]);
  }
  sort(crr.begin(), crr.end());
  vector<bool> drr((int)((crr).size()), false);
  map<pair<long long int, long long int>, long long int> mp;
  ;
  for (int i = 0; i < (int)((crr).size()); i++) {
    if (drr[i]) continue;
    long long int r = crr[i].second;
    ;
    for (int j = i + 1; j < (int)((crr).size()); j++) {
      if (r >= crr[j].first) {
        r = max(crr[j].second, r);
        drr[j] = true;
      } else {
        break;
      }
    }
    drr[i] = true;
    mp[make_pair(crr[i].first, r)] = 1;
  }
  long long int sum = 0;
  bool valid = false;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    if (x <= it->first.second && x >= it->first.first) valid = true;
    sum += it->first.second - it->first.first + 1;
  }
  if (!valid) sum++;
  cout << sum << "\n";
}
int32_t main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> tes;
  while (tes--) {
    solve();
  }
}
