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 boolean[] visited;
	static Stack<Integer> stack;
	static ArrayList<Integer>[] adjList;

	public static void topoSort(int cur)
	{
		visited[cur] = true;
		for(int nxt : adjList[cur])
			if(!visited[nxt])
				topoSort(nxt);
		stack.push(cur);
	}
	
	public static void dfs(int cur)
	{
		visited[cur] = true;
		for(int nxt : adjList[cur])
			if(!visited[nxt])
				dfs(nxt);
	}

	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int capital = sc.nextInt() - 1;
		
		stack = new Stack<>();
		visited = new boolean[n];
		adjList = new ArrayList[n];
		for(int i = 0; i < n; i++)
			adjList[i] = new ArrayList<>();
		
		for(int i = 0; i < m; i++)
			adjList[sc.nextInt() - 1].add(sc.nextInt() - 1);
		
		dfs(capital);
		boolean[] reach = visited.clone();
		visited = new boolean[n];
		for(int i  = 0; i < n; i++)
			if(!visited[i])
				topoSort(i);
		
		int count = 0;
		visited = new boolean[n];
		while(!stack.isEmpty())
		{
			int cur = stack.pop();
			if(!visited[cur] && !reach[cur])
			{
				adjList[capital].add(cur);
				dfs(capital);
				count++;
			}
		}
		
		System.out.println(count);
		
	}

	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();}
	
	
	}

}
