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.ArrayList;
import java.util.Stack;
import java.util.StringTokenizer;

public class E {

    static ArrayList<Integer>[] g1, g2;
    static int[] sccIndex;
    static boolean[] vis;
    static Stack<Integer> stack;

    static void dfs1(int u) {
        vis[u] = true;
        for(int v: g1[u]) if(!vis[v])
            dfs1(v);
        stack.push(u);
    }

    static void dfs2(int u, int idx) {
        vis[u] = true;
        sccIndex[u] = idx;
        for(int v: g2[u]) if(!vis[v])
            dfs2(v, idx);
    }

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

        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = sc.nextInt(), m = sc.nextInt(), s = sc.nextInt() - 1;
        g1 = new ArrayList[n];
        g2 = new ArrayList[n];
        for(int i = 0; i < n; ++i) {
            g1[i] = new ArrayList<>();
            g2[i] = new ArrayList<>();
        }
        while(m-->0) {
            int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
            g1[u].add(v);
            g2[v].add(u);
        }
        stack = new Stack<>();
        vis = new boolean[n];
        for(int i = 0; i < n; ++i) if(!vis[i])
            dfs1(i);
        vis = new boolean[n];
        sccIndex = new int[n];
        int idx = 0;
        while(!stack.isEmpty()) {
            int i = stack.pop();
            if(!vis[i])
                dfs2(i, idx++);
        }
        int[] deg = new int[idx];
        for(int i = 0; i < n; ++i)
            for(int v: g1[i])
                if(sccIndex[i] != sccIndex[v])
                    deg[sccIndex[v]]++;
        int ans = 0;
        for(int i = 0, k = sccIndex[s]; i < idx; ++i)
            if(i != k && deg[i] == 0)
                ++ans;
        out.println(ans);
        out.close();
    }

    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {return Integer.parseInt(next());}

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public double nextDouble() throws IOException { return Double.parseDouble(next()); }

        public String nextLine() throws IOException {return br.readLine();}

        public boolean ready() throws IOException { return br.ready(); }

    }
}
