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.io.*;
import java.util.*;
public class CF999E {
    static int count = 0;
    public static void main(String args[]) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] s = br.readLine().split(" ");
        int n = Integer.parseInt(s[0]);
        int m = Integer.parseInt(s[1]);
        int c = Integer.parseInt(s[2]);
        c--;

        Node[] nodes = new Node[n];
        for(int i=0;i<n;i++){
            nodes[i] = new Node(i);
        }
        for(int i=0;i<m;i++){
            s = br.readLine().split(" ");
            int u = Integer.parseInt(s[0]);
            int v = Integer.parseInt(s[1]);
            u--;
            v--;
            nodes[u].conn.add(nodes[v]);
        }
        int ans = 0;
        visitNode(nodes[c]);

        for(Node node: nodes){
            if(!node.visited)
            {
                count = 0;
                visitNode1(node);
                for(Node n1: nodes){
                    n1.visited1 = false;
                }
                node.count = count;
            }
        }

        Arrays.sort(nodes, new NodeSort());
        /*for(Node n1: nodes){
            if(!n1.visited){
                System.out.println((n1.index+1)+": "+n1.count);
            }
        }*/
        while(true){
            Node node = null;
            for(Node n1: nodes){
                if(!n1.visited){
                    node = n1;
                    break;
                }
            }
            if(node == null){
                break;
            }
            //System.out.println("Edge Added;"+(node.index+1));
            nodes[c].conn.add(node);
            visitNode(node);
            ans++;
        }
        System.out.println(ans);

    }
    private static void visitNode(Node node){
        node.visited = true;
        for(Node n1: node.conn){
            if(!n1.visited)
            visitNode(n1);
        }
    }

    private static void visitNode1(Node node){
        node.visited1 = true;
        count++;
        for(Node n1: node.conn){
            if(n1.visited){
                continue;
            }
            if(!n1.visited1)
            visitNode1(n1);
        }
    }

}
class Node{
    int index;
    ArrayList<Node> conn;
    boolean visited;
    boolean visited1;
    int count = 0;
    public Node(int index){
        this.index = index;
        conn = new ArrayList<Node>();
    }
}
class NodeSort implements Comparator<Node>{
    public int compare(Node n1, Node n2){
        return n2.count - n1.count;
    }
}
