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.
import java.util.*;
import java.io.*;
public class ReachabilityFromCapital {
  static int visit [];
  static ArrayList<Integer> e[] = new ArrayList[10000];
  static void dfs(int s, int p)
  {
    visit[s] = p;
    for (int i = 0;i < e[s].size();++i) {
      int v = e[s].get(i);
      if (visit[v] != p) dfs(v, p);
    }
  }
  public static void main(String[] args) {
      visit = new int[10000];
      int n;
      int m;
      int s;
      
      Scanner in = new Scanner(System.in);

      n = in.nextInt();
      m = in.nextInt();
      s = in.nextInt();

      for (int i = 0; i <= n; i++) {
        e[i] = new ArrayList<>();
      }
      for (int i = 0;i < m;++i) {
        int u = in.nextInt();
        int v = in.nextInt();
        e[u].add(v);
      }
      
      for (int i = 1;i <= n;++i) {
        if (visit[i] == 0) {
          dfs(i, i);
        }
      }
      dfs(s, s);
      
      Set edges = new HashSet<>();
      for (int i = 1; i <= n; i++) {
        edges.add(visit[i]);
      }
      System.out.println(edges.size() - 1); 
  }
}




