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.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Main {

    static boolean[] visited;
    static ArrayList<Integer>[] adj;

    public static void main(String[] args) throws Exception {

        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int n, m, s;
        String[] temp = in.readLine().split(" ");
        n = Integer.parseInt(temp[0]);
        m = Integer.parseInt(temp[1]);
        s = Integer.parseInt(temp[2]) - 1;

        adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }

        for (int i = 0; i < m; i++) {
            temp = in.readLine().split(" ");
            int a = Integer.parseInt(temp[0]) - 1;
            int b = Integer.parseInt(temp[1]) - 1;
            adj[a].add(b);
        }

        int[] sizes = new int[n];
        for (int i = 0; i < n; i++) {
            visited = new boolean[n];
            sizes[i] = dfs(i);
        }
        ArrayList<Integer> byCount = new ArrayList<>();
        for (int i = 0; i < n; i++)
            byCount.add(i);
        byCount.sort(new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return Integer.compare(sizes[o2], sizes[o1]);
            }
        });

        visited = new boolean[n];
        int comp = 0;
        dfs(s);
        comp++;
        for (int i: byCount) {
            if (!visited[i]) {
                dfs(i);
                comp++;
            }
        }

        if (comp == 1) {
            System.out.println(0);
        } else {
            System.out.println(comp - 1);
        }

    }

    private static int dfs(int i) {
        int n = 1;
        visited[i] = true;
        for (int c: adj[i]) {
            if (!visited[c]) {
                n += dfs(c);
            }
        }
        return n;
    }

}