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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') f = -1, c = getchar();
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return f * x;
}
const long long N = 2e5 + 7;
const long long M = 5e4 + 7;
const long long MAXN = 4e18 + 7;
const long long Mod = 1e9 + 7;
long long _, i, j, k, n, m, p, s, T, l, r, o, u, v, w, x, y, z, st, en, sx, sy,
    ans, th, ma, mi, mod, sum, num, len, res, flag, cas, ret, mid, cnt, now,
    nex, tmp;
long long c[N];
char ch;
queue<long long> q;
vector<long long> g[N];
string s1, s2, s3;
struct node {
  long long l, r;
} a[N];
bool cmp(node a, node b) {
  long long x1 = min(abs(a.l - x), abs(a.r - x));
  long long x2 = min(abs(b.l - x), abs(b.r - x));
  return x1 < x2;
}
signed main() {
  ios::sync_with_stdio(false);
  long long T = 1;
  double x1, x2, x3, y1, y2, y3, z1, z2, z3;
  cin >> T;
  while (T--) {
    cin >> n >> x >> m;
    l = r = x;
    cnt = 0;
    for (i = 1; i <= m; i++) {
      cin >> a[i].l >> a[i].r;
      if (a[i].l <= l && r <= a[i].r) l = a[i].l, r = a[i].r;
      if (l <= a[i].l && a[i].l <= r) r = max(r, a[i].r);
      if (l <= a[i].r && a[i].r <= r) l = min(l, a[i].l);
    }
    cout << r - l + 1 << endl;
  }
  return 0;
}
