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 n, m, s;
vector<long long> v[5005];
bool vis[5005];
bool vis2[5005];
vector<pair<long long, long long> > p;
set<long long> se;
long long sum;
void dfs(int s) {
  sum++;
  vis[s] = 1;
  for (int i = 0; i < v[s].size(); i++) {
    if (!vis[v[s][i]]) {
      dfs(v[s][i]);
    }
  }
}
int main() {
  long long cnt = 0;
  cin >> n >> m >> s;
  for (int i = 1; i <= n; i++) {
    se.insert(i);
  }
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    v[x].push_back(y);
    if (se.count(y)) {
      se.erase(y);
    }
  }
  dfs(s);
  cnt++;
  if (se.count(s)) {
    se.erase(s);
  }
  set<long long>::iterator it = se.begin();
  for (it; it != se.end(); it++) {
    if (!vis[*it]) {
      dfs(*it);
      cnt++;
    }
  }
  for (int i = 1; i <= n; i++) {
    vis2[i] = vis[i];
  }
  for (int i = 1; i <= n; i++) {
    if (!vis2[i]) {
      sum = 0;
      dfs(i);
      memset(vis, 0, sizeof vis);
      p.push_back({sum, i});
    }
  }
  sort(p.begin(), p.end());
  for (int i = 1; i <= n; i++) {
    vis[i] = vis2[i];
  }
  for (int i = p.size() - 1; i >= 0; i--) {
    if (!vis[p[i].second]) {
      cnt++;
      dfs(p[i].second);
    }
  }
  cout << cnt - 1 << endl;
  return 0;
}
