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;
int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
const long long INF = 1LL << 60;
const long long MOD = 1e9 + 7;
bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s) {
  return f.second > s.second;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long conbinationMemo[100][100];
void cmemoInit() {
  for (int i = 0; i < (int)(100); i++) {
    for (int j = 0; j < (int)(100); j++) {
      conbinationMemo[i][j] = -1;
    }
  }
}
long long nCr(long long n, long long r) {
  if (conbinationMemo[n][r] != -1) return conbinationMemo[n][r];
  if (r == 0 || r == n) {
    return 1;
  } else if (r == 1) {
    return n;
  }
  return conbinationMemo[n][r] = (nCr(n - 1, r) + nCr(n - 1, r - 1));
}
long long nPr(long long n, long long r) {
  r = n - r;
  long long ret = 1;
  for (long long i = n; i >= r + 1; i--) ret *= i;
  return ret;
}
int main(void) {
  long long t;
  cin >> t;
  for (int ii = 0; ii < (int)(t); ii++) {
    long long n, m, x;
    cin >> n >> x >> m;
    long long nowr = x;
    long long nowl = x;
    for (int i = 0; i < (int)(m); i++) {
      long long l, r;
      cin >> l >> r;
      if (nowr < l || nowl > r) continue;
      nowr = max(nowr, r);
      nowl = min(nowl, l);
    }
    cout << nowr - nowl + 1 << endl;
  }
}
