There are n cities and m roads in Berland. Each road connects a pair of cities. The roads in Berland are one-way.

What is the minimum number of new roads that need to be built to make all the cities reachable from the capital?

New roads will also be one-way.

Input

The first line of input consists of three integers n, m and s (1 ≤ n ≤ 5000, 0 ≤ m ≤ 5000, 1 ≤ s ≤ n) — the number of cities, the number of roads and the index of the capital. Cities are indexed from 1 to n.

The following m lines contain roads: road i is given as a pair of cities u_i, v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i). For each pair of cities (u, v), there can be at most one road from u to v. Roads in opposite directions between a pair of cities are allowed (i.e. from u to v and from v to u).

Output

Print one integer — the minimum number of extra roads needed to make all the cities reachable from city s. If all the cities are already reachable from s, print 0.

Examples

Input

9 9 1
1 2
1 3
2 3
1 5
5 6
6 1
1 8
9 8
7 1


Output

3


Input

5 4 5
1 2
2 3
3 4
4 1


Output

1

Note

The first example is illustrated by the following:

<image>

For example, you can add roads (6, 4), (7, 9), (1, 7) to make all the cities reachable from s = 1.

The second example is illustrated by the following:

<image>

In this example, you can add any one of the roads (5, 1), (5, 2), (5, 3), (5, 4) to make all the cities reachable from s = 5.
#include <bits/stdc++.h>
using namespace std;
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<long long> v[5001];
set<long long> s;
long long n, m, cap;
vector<bool> vis;
void dfs(long long cur) {
  vis[cur] = true;
  for (long long adj : v[cur]) {
    if (!vis[adj]) {
      dfs(adj);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> m >> cap;
  long long x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    v[x].push_back(y);
  }
  vis = vector<bool>(n + 1, false);
  dfs(cap);
  long long count = 0;
  long long flag = 0;
  set<long long>::iterator it;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      s.insert(i);
    }
  }
  for (int i = 1; i <= n; i++) {
    if (s.find(i) != s.end()) {
      vis = vector<bool>(n + 1, false);
      dfs(i);
      for (int j = 1; j <= n; j++) {
        if (vis[j] == 1 && i != j && s.find(j) != s.end()) {
          s.erase(j);
        }
      }
    }
  }
  cout << s.size() << endl;
}
