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<bool> mark;
bool used[5010];
vector<vector<int>> adjacenciesOf;
void dfs(int v) {
  stack<int> s;
  s.push(v);
  while (!s.empty()) {
    int vertex = s.top();
    s.pop();
    mark[vertex] = true;
    for (int adj : adjacenciesOf[vertex]) {
      if (!mark[adj]) s.push(adj);
    }
  }
}
int dfs1(int v) {
  int cnt = 0;
  stack<int> s;
  s.push(v);
  while (!s.empty()) {
    int vertex = s.top();
    s.pop();
    used[vertex] = true;
    ++cnt;
    for (int adj : adjacenciesOf[vertex]) {
      if (!used[adj] && !mark[adj]) s.push(adj);
    }
  }
  return cnt;
}
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  mark.resize(n + 1, false);
  adjacenciesOf.resize(n + 1);
  while (m--) {
    int a, b;
    cin >> a >> b;
    adjacenciesOf[a].push_back(b);
  }
  dfs(s);
  vector<pair<int, int>> val;
  for (int i = 1; i <= n; ++i) {
    if (!mark[i]) {
      memset(used, false, sizeof(used));
      val.emplace_back(make_pair(dfs1(i), i));
    }
  }
  sort(val.begin(), val.end());
  reverse(val.begin(), val.end());
  int ans = 0;
  for (pair<int, int> it : val) {
    if (!mark[it.second]) {
      ++ans;
      dfs(it.second);
    }
  }
  cout << ans << endl;
  return 0;
}
