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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;

public class Ex05_sol2 {

    private static ArrayList<Integer> adjList[];
    private static boolean[] visited, reachable;
    private static HashSet<Integer> r;
    private static boolean[] d2;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int s = sc.nextInt() - 1;
        adjList = new ArrayList[n];
        reachable = new boolean[n];
        d2 = new boolean[n];
        r = new HashSet<>();

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

        for (int i = 0; i < m; i++) {
            int u = sc.nextInt() - 1;
            int v = sc.nextInt() - 1;
            adjList[u].add(v);
        }
        dfs(s);

        for (int i = 0; i < n; i++) {
            if (!reachable[i]) {
                visited = new boolean[n];
                r.add(i);
                dfs2(i);
                d2[i] = true;
            }
        }
        System.out.println(r.size());
    }

    private static void dfs2(int u) {
        visited[u] = true;
        reachable[u] = true;
        for (int v : adjList[u]) {
            if (d2[v])
                r.remove(v);
            if (!visited[v])
                dfs2(v);
        }
    }

    static void dfs(int u) {
        reachable[u] = true;
        for (int v : adjList[u]) {
            if (!reachable[v])
                dfs(v);
        }
    }
}
