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 + 5;
int T, n, x, m, a[N], b[N], cnta, le[N], ri[N], y[505];
map<int, int> vis, to;
int main() {
  scanf("%d", &T);
  while (T--) {
    memset(y, 0, sizeof(y));
    scanf("%d%d%d", &n, &x, &m);
    to.clear();
    cnta = 0;
    a[++cnta] = x;
    for (long long i = 1; i <= m; i++) {
      scanf("%d%d", &le[i], &ri[i]);
      a[++cnta] = le[i];
      a[++cnta] = ri[i];
    }
    sort(a + 1, a + cnta + 1);
    int len = unique(a + 1, a + cnta + 1) - a - 1;
    for (long long i = 1; i <= len; i++) to[a[i]] = i;
    y[to[x]] = 1;
    int flg = 0;
    for (long long i = 1; i <= m; i++) {
      int fg = 0;
      for (long long j = to[le[i]]; j <= to[ri[i]]; j++) {
        if (y[j]) fg = 1;
      }
      if (fg) {
        if (le[i] != ri[i]) flg = 1;
        for (long long j = to[le[i]]; j <= to[ri[i]]; j++) {
          y[j] = 1;
        }
      }
    }
    a[0] = a[1];
    int ans = 0, f1 = 0;
    for (long long i = 1; i <= len; i++) {
      if (y[i]) {
        if (f1)
          ans += a[i] - a[i - 1];
        else
          f1 = 1;
      }
    }
    if (flg)
      printf("%d\n", ans + 1);
    else
      printf("1\n");
  }
}
