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;
vector<int> g[10000];
int vis1[10000], vis2[10000];
int cnt;
void dfs(int node) {
  if (vis1[node]) return;
  vis1[node] = 1;
  for (auto i : g[node]) {
    if (vis1[i] == 0) dfs(i);
  }
}
void dfs1(int node) {
  if (vis2[node]) return;
  vis2[node] = 1;
  for (auto i : g[node]) {
    if (vis2[i] == 0) dfs1(i);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, c;
  cin >> n >> m >> c;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
  }
  dfs(c);
  vector<pair<int, int> > s;
  for (int i = 1; i <= n; i++) {
    cnt = 0;
    if (!vis1[i]) {
      memset(vis2, 0, sizeof(vis2));
      dfs1(i);
      for (int j = 1; j <= n; j++) {
        if (vis2[j] == 1 and vis1[j] == 0) cnt++;
      }
      s.push_back({cnt, i});
    }
  }
  sort(s.begin(), s.end());
  reverse(s.begin(), s.end());
  int ans = 0;
  for (auto i : s) {
    memset(vis2, 0, sizeof(vis2));
    cnt = 0;
    if (vis1[i.second] == 0) {
      dfs(i.second);
      ans++;
    }
  }
  cout << ans << endl;
  cerr << "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << "ms"
       << '\n';
  return 0;
}
