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 comp(pair<long long, long long> a, pair<long long, long long> b) {
  if (a.first != b.first)
    return a.first < b.first;
  else
    return a.second > b.second;
}
long long ex(long long n, long long ans) {
  if (n == 1)
    return ans;
  else if (n % 2)
    return (ans % ((1000000007)) * ex(n / 2, (ans * ans) % (1000000007))) %
           (1000000007);
  else
    return ex(n / 2, (ans * ans) % (1000000007));
}
long long kmp(string x, long long m) {
  vector<long long> pi(m, 0);
  pi[0] = 0;
  long long j = 0;
  for (long long i = 1; i < m; i++) {
    j = pi[i - 1];
    while (j > 0 && x[i] != x[j]) j = pi[j - 1];
    if (x[j] == x[i]) j++;
    pi[i] = j;
  }
  return j;
}
string kmp1(string x, long long m) {
  vector<long long> pi(m, 0);
  pi[0] = 0;
  long long ma = 0;
  long long j = 0;
  long long pos = 0;
  for (long long i = 1; i < m; i++) {
    j = pi[i - 1];
    while (j > 0 && x[i] != x[j]) j = pi[j - 1];
    if (x[j] == x[i]) j++;
    pi[i] = j;
    if (x[i] == '#') pos = i;
  }
  long long p = pos;
  for (long long i = (pos); i < (m); ++i) {
    if (pi[i] > ma) {
      ma = pi[i];
      p = i;
    }
  }
  return x.substr(p - ma + 1, ma);
}
void dfs(vector<long long> adj[], long long fl[], long long i, long long& cn) {
  fl[i] = 1;
  cn++;
  for (long long j = (0); j < (adj[i].size()); ++j) {
    if (!fl[adj[i][j]]) {
      dfs(adj, fl, adj[i][j], cn);
    }
  }
}
int isp(long long x) {
  for (long long i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
struct Compare {
  bool operator()(long long const& p1, long long const& p2) {
    if (p1 < p2) {
      return true;
    }
    return false;
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t1 = 1;
  cin >> t1;
  while (t1--) {
    long long n, m, x;
    cin >> n >> x >> m;
    long long l = x, r = x;
    while (m--) {
      long long a, b;
      cin >> a >> b;
      if ((a <= l && l <= b) || (r <= b && r >= a))
        l = min(l, a), r = max(b, r);
    }
    cout << (r - l + 1) << " ";
    cout << "\n";
  }
}
