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.

// Imports
import java.util.*;
import java.io.*;

public class E999 {
    
    /**
     * @param args the command line arguments
     * @throws IOException, FileNotFoundException 
     */
    public static void main(String[] args) throws IOException, FileNotFoundException {
        
        // TODO UNCOMMENT WHEN ALGORITHM CORRECT
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        
        
        // TODO code application logic here
        StringTokenizer st = new StringTokenizer(f.readLine());
        int N = Integer.parseInt(st.nextToken());
        int M = Integer.parseInt(st.nextToken());
        int S = Integer.parseInt(st.nextToken()) - 1;
        
        ArrayList<Integer>[] adj = new ArrayList[N];
        boolean[] isGood = new boolean[N];
        int[][] nextTo = new int[N][2];
        
        for(int i = 0; i < N; i++) {
            adj[i] = new ArrayList<>();
            nextTo[i][1] = i;
        }
        
        for(int i = 0; i < M; i++) {
            StringTokenizer edge = new StringTokenizer(f.readLine());
            int A = Integer.parseInt(edge.nextToken()) - 1;
            int B = Integer.parseInt(edge.nextToken()) - 1;
            adj[A].add(B);
        }
        
        boolean[] visited = new boolean[N];
        
        Queue<Integer> bfs = new ArrayDeque<>();
        bfs.add(S);
        visited[S] = true;
        isGood[S] = true;
        while(!bfs.isEmpty()) {
            int curr = bfs.poll();
            
            for(int i : adj[curr]) {
                if(!visited[i]) {
                    visited[i] = true;
                    isGood[i] = true;
                    nextTo[i][0] = -1;
                    bfs.add(i);
                }
            }
        }
        
        for(int i = 0; i < N; i++) {
            visited = new boolean[N];
            if(!isGood[i]) {
                int sum = 0;
                visited[i] = true;
                bfs.add(i);
                
                while(!bfs.isEmpty()) {
                    int curr = bfs.poll();
                    sum++;
                    
                    for(int j : adj[curr]) {
                        if(!visited[j] && !isGood[j]) {
                            visited[j] = true;
                            bfs.add(j);
                        }
                    }
                }
                
                nextTo[i][0] = sum;
            }
        }
        
        Arrays.sort(nextTo, (int[] o1, int[] o2) -> o2[0] - o1[0]);
        
        int curr = 0;
        for(int i = 0; i < nextTo.length; i++) {
            if(nextTo[i][0] == -1) {
                break;
            }
            
            if(!isGood[nextTo[i][1]]) {
                curr++;
                visited = new boolean[N];
                visited[nextTo[i][1]] = true;
                isGood[nextTo[i][1]] = true;
                bfs.add(nextTo[i][1]);
                
                while(!bfs.isEmpty()) {
                    int next = bfs.poll();
                    
                    for(int j : adj[next]) {
                        if(!visited[j]) {
                            visited[j] = true;
                            isGood[j] = true;
                            bfs.add(j);
                        }
                    }
                }
            }
        }
        
        System.out.println(curr);
    }

}
