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 INF = 1e9;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
long long power(long long a, long long b, long long m = MOD) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return ans;
}
long long dir[] = {-1, 0, 1, 0, -1};
long long dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long N = 1e5 + 1;
bool check(long long l, long long h, long long x, long long y) {
  if (l > y or h < x) return false;
  return true;
}
void solve() {
  long long i, n, m, x;
  cin >> n >> x >> m;
  set<pair<long long, long long> > st;
  st.insert({x, x});
  for (i = 0; i < m; i++) {
    long long l, h;
    cin >> l >> h;
    for (auto x : st) {
      if (check(l, h, x.first, x.second)) {
        st.erase(x);
        l = min(l, x.first);
        h = max(x.second, h);
        st.insert({l, h});
      }
    }
  }
  long long l, h;
  l = (*st.begin()).first;
  h = (*st.begin()).second;
  cout << h - l + 1 << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  cin >> t;
  while (t--) solve();
}
