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<long long int> vect[5001];
bool visited[5001] = {false};
bool visited1[5001] = {false};
long long int x = 0;
void dfs(long long int s1) {
  visited1[s1] = true;
  x++;
  for (long long int i = 0; i < vect[s1].size(); i++) {
    if (!visited1[vect[s1][i]]) dfs(vect[s1][i]);
  }
}
void dfs1(long long int s1) {
  visited[s1] = true;
  for (long long int i = 0; i < vect[s1].size(); i++) {
    if (!visited[vect[s1][i]]) dfs1(vect[s1][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  long long int n, m, s;
  cin >> n >> m >> s;
  long long int i, a, b;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    vect[a].push_back(b);
  }
  long long int j, c = 0;
  for (j = 0; j < vect[s].size(); j++) dfs1(vect[s][j]);
  vector<pair<long long int, long long int> > vect1;
  for (i = 1; i <= n; i++) {
    if (!visited[i] && i != s) {
      x = 0;
      memset(visited1, false, sizeof(visited1));
      dfs(i);
      vect1.push_back(make_pair(x, i));
    }
  }
  sort(vect1.begin(), vect1.end());
  reverse(vect1.begin(), vect1.end());
  for (i = 0; i < vect1.size(); i++) {
    if (!visited[vect1[i].second]) {
      c++;
      dfs1(vect1[i].second);
    }
  }
  cout << c;
  return 0;
}
