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>
int n, m;
using namespace std;
vector<int> used(5005, 0);
vector<int> used2(5005, 0);
int jlo1 = 3;
vector<int> a;
long long res = 0;
vector<vector<int>> graph(5005, a);
vector<int> b;
vector<vector<int>> graph1(5005, b);
int q = 0;
bool flag = false;
void dfs(int p) {
  {
    q++;
    used[p] = 1;
    for (int i = 0; i < graph[p].size(); i++) {
      if (!used[graph[p][i]]) dfs(graph[p][i]);
    }
  }
}
void dfs2(int p) {
  {
    q++;
    used2[p] = 1;
    for (int i = 0; i < graph[p].size(); i++) {
      if (!used2[graph[p][i]]) dfs2(graph[p][i]);
    }
  }
}
int main() {
  int k, t, t1;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> t >> t1;
    graph[t].push_back(t1);
  }
  dfs(k);
  vector<pair<int, int>> p_q;
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      used2.assign(5005, 0);
      q = 0;
      dfs2(i);
      p_q.push_back(make_pair(q, i));
    }
  }
  sort(p_q.rbegin(), p_q.rend());
  long long res = 0;
  for (int i = 0; i < p_q.size(); i++) {
    if (!used[p_q[i].second]) {
      res++;
      dfs(p_q[i].second);
    }
  }
  cout << res << "\n";
  return 0;
}
