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> visited;
vector<vector<int> > adj;
vector<int> unvisited;
vector<int> unvisited_visited;
void print_v(vector<int> v) {
  for (auto it : v) {
    cout << it << ",";
  }
  cout << endl;
}
void dfs(int s) {
  visited[s] = 1;
  for (int i = 0; i < adj[s].size(); ++i) {
    if (visited[adj[s][i]] != 1) dfs(adj[s][i]);
  }
}
void dfs_u(int s, int n) {
  unvisited_visited[s] = n;
  for (int i = 0; i < adj[s].size(); ++i) {
    if (unvisited_visited[adj[s][i]] != n) dfs_u(adj[s][i], n);
  }
}
void find_connected_components() {
  int n = 1;
  for (int i = 0; i < visited.size(); ++i) {
    if (unvisited_visited[i] == 0 && visited[i] == 0) {
      dfs_u(i, n);
      n++;
    }
  }
}
int count_unconnected_components() {
  int count = 0;
  for (int i = 0; i < unvisited_visited.size(); ++i) {
    int temp = unvisited_visited[i];
    if (unvisited_visited[i] > 0) {
      for (int j = i; j < unvisited_visited.size(); ++j) {
        if (unvisited_visited[j] == temp)
          unvisited_visited[j] = -unvisited_visited[j];
      }
      count++;
    }
  }
  return count;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  s--;
  visited.resize(n);
  adj.resize(n);
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
  }
  dfs(s);
  unvisited_visited = visited;
  for (int i = 0; i < n; ++i) {
    unvisited_visited[i] = 0;
  }
  find_connected_components();
  cout << count_unconnected_components() << endl;
  return 0;
}
