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;
typedef set<int>* edge_map;
void df(int x, edge_map edges, bool* visited) {
  visited[x] = true;
  set<int> nodes = edges[x];
  for (set<int>::iterator node = nodes.begin(); node != nodes.end(); ++node) {
    if (!visited[*node]) {
      df(*node, edges, visited);
    }
  }
}
void calc_top_order(int x, edge_map edges, bool* visited,
                    list<int>& ordered_nodes) {
  visited[x] = true;
  set<int> nodes = edges[x];
  for (set<int>::iterator node = nodes.begin(); node != nodes.end(); ++node) {
    if (!visited[*node]) {
      calc_top_order(*node, edges, visited, ordered_nodes);
    }
  }
  ordered_nodes.push_front(x);
}
int main() {
  istream* is = &cin;
  int n, m, s;
  *is >> n >> m >> s;
  s--;
  edge_map edges = new set<int>[n];
  for (int i = 0; i < n; i++) edges[i] = set<int>();
  for (int i = 0; i < m; i++) {
    int u, v;
    *is >> u >> v;
    edges[u - 1].insert(v - 1);
  }
  bool* visited = new bool[n];
  memset(visited, false, n * sizeof(bool));
  df(s, edges, visited);
  bool* temp_visited = new bool[n];
  memcpy(temp_visited, visited, n * sizeof(bool));
  int cnt = 0;
  list<int> ordered_nodes;
  for (int i = 0; i < n; i++) {
    if (!visited[i] && !temp_visited[i])
      calc_top_order(i, edges, temp_visited, ordered_nodes);
  }
  while (!ordered_nodes.empty()) {
    int node = ordered_nodes.front();
    ordered_nodes.pop_front();
    if (!visited[node]) {
      cnt++;
      df(node, edges, visited);
    }
  }
  cout << cnt;
  return 0;
}
