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.*;
import java.io.*;
public class A{
	static LinkedList<Integer> list[];
	static boolean visited[];
	static Stack<Integer> stack = new Stack<Integer>();
	static void dfs(int u,boolean p)
	{
		if(p)
		{
			if(visited[u])
				return;
			visited[u] = true;
		}
		else
		{
			if(!visited[u])
				return;
			visited[u] = false;
		}
		for(Integer v : list[u])
			dfs(v,p);
		if(p)
			stack.push(u);
	}
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int m = scan.nextInt();
		int start = scan.nextInt();
		list = new LinkedList[n + 1];
		visited = new boolean[n + 1];
		for(int i = 0;i <=n;i++)
			list[i] = new LinkedList<>();
		for(int i = 0;i < m;i++)
		{
			int u = scan.nextInt();
			int v = scan.nextInt();
			list[u].add(v);
		}
		for(int i = 1;i <= n;i++)
			dfs(i,true);
		dfs(start,false);
		int count = 0;
		while(!stack.isEmpty())
		{
			int u = stack.pop();
			if(!visited[u])
				continue;
			else
			{
				dfs(u,false);
				count++;
			}
		}
		System.out.println(count);
	}
}