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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.*;
import java.util.*;

public class D {
	static int n, m;
	static int[] vis;
	static boolean reach[];
	static int vid;
	static int[] inDeg;

	static void dfs(int u) {
		reach[u] = true;
		vis[u] = vid;
		for (int v : adj[u])
			if (vis[v] != vid)
				dfs(v);
	}

	static ArrayList<Integer>[] adj;
	static int V, counter, SCC, dfs_num[], dfs_low[];
	static int[] inSCC;
	static Stack<Integer> stack;
	static int cap;

	static void tarjanSCC() // O(V + E)
	{
		dfs_num = new int[V];
		dfs_low = new int[V];
		inSCC = new int[V];
		stack = new Stack<>();
		for (int i = 0; i < V; ++i)
			if (!reach[i] && dfs_num[i] == 0)
				tarjanSCC(i);
	}

	static void tarjanSCC(int u) {
		dfs_num[u] = dfs_low[u] = ++counter;
		stack.push(u);
		for (int v : adj[u])
			if (!reach[v]) {
				if (dfs_num[v] == 0)
					tarjanSCC(v);
				if (0 == inSCC[v])
					dfs_low[u] = Math.min(dfs_low[u], dfs_low[v]);
			}
		if (dfs_num[u] == dfs_low[u]) {
			SCC++;
			while (true) {
				int v = stack.pop();
				inSCC[v] = SCCid;
				if (v == u)
					break;
			}
			SCCid++;
		}
	}

	static int SCCid;

	public static void main(String[] args) throws Throwable {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		n = sc.nextInt();
		m = sc.nextInt();
		adj = new ArrayList[n];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList<>();
		cap = sc.nextInt() - 1;
		inDeg = new int[n];
		for (int i = 0; i < m; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			adj[u].add(v);
		}
		int ans = 0;
		vis = new int[n];
		reach = new boolean[n];
		SCCid = 1;
		vid = 1;
		V = n;
		dfs(cap);
		tarjanSCC();
		for (int i = 0; i < n; i++) {
			if (!reach[i])
				for (int v : adj[i])
					if (!reach[v]) {
						if (inSCC[i] != inSCC[v])
							inDeg[inSCC[v]]++;
					}
		}
		for (int i = 1; i <= SCC; i++)
			if (inDeg[i] == 0)
				ans++;
		pw.println(ans);
		pw.flush();
		pw.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 String nextLine() throws IOException {
			return br.readLine();
		}

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

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