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 E {


	static int V, E, dfs_counter, connected;
	static ArrayList<Integer>[] adjList;
	static Stack<Integer> stack;

	

	static boolean[] vis;

	static void dfs1(int u) {

		vis[u] = true;

		for (int v : adjList[u])
			if (!vis[v])
				dfs1(v);
		
		stack.add(u);

	}
	
	static void dfs(int u) {

		vis[u] = true;

		for (int v : adjList[u])
			if (!vis[v])
				dfs(v);
		

	}

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

		Scanner sc = new Scanner(System.in);

		V = sc.nextInt();
		E = sc.nextInt();
		int s = sc.nextInt() - 1;

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

		while (E-- > 0) {

			int u = sc.nextInt();
			int v = sc.nextInt();
			adjList[u - 1].add(v - 1);
		}

		int ans = 0;
		
		stack = new Stack<>();
		
		for(int i = 0 ; i < V ; i++)
			if(!vis[i])
				dfs1(i);
		
		

		while(!stack.isEmpty())
		{
			int u = stack.pop();
			
			if(u == s) continue;
			
			vis = new boolean[V];
			
			dfs(s);
			
			if(!vis[u]) {
				adjList[s].add(u);
				ans++;
			}
		}
		

		System.out.println(ans);

	}

	static void shuffle(long[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			long tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}

	static void shuffle(int[] a) {
		int n = a.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			int tmp = a[i];
			a[i] = a[r];
			a[r] = tmp;
		}
	}

	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 {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

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

	}

}