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{
	public static void main(String[] args)throws Throwable {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out);
		
		V=n=sc.nextInt();
		int m=sc.nextInt();
		int s=sc.nextInt()-1;
		adj=new ArrayList[n];
		adjList=new ArrayList[n];
		for(int i=0;i<n;i++) {
			adj[i]=new ArrayList<>();
			adjList[i]=new ArrayList<>();			
		}
		while(m-->0) {
			int u=sc.nextInt()-1;
			int v=sc.nextInt()-1;
			adj[u].add(v);
		}
		vis=new boolean [n];
		dfs(s);
		for(int i=0;i<n;i++) {
			if(vis[i])
				continue;
			for(int v : adj[i])
				if(!vis[v])
					adjList[i].add(v);
		}
		tarjanSCC();
		int [] deg=new int [n];
		int ans=-cntV;
		for(int i=0;i<n;i++)
			for(int v : adjList[i])
				if(comp[v]!=comp[i])
					deg[comp[v]]++;
		for(int i=0;i<SCC;i++)
			if(deg[i]==0)
				ans++;
		
		pw.println(ans);
		pw.flush();
		pw.close();
    }
	
	static int n,cntV;
	static ArrayList<Integer> [] adj;
	static boolean [] vis;
	
	static void dfs(int u) {
		cntV++;
		vis[u]=true;
		for(int v : adj[u])
			if(!vis[v])
				dfs(v);
	}
	
	static ArrayList<Integer>[] adjList;
	static int V, counter, SCC, dfs_num[], dfs_low[],comp[];
	static boolean[] inSCC;
	static Stack<Integer> stack;
		
	
	static void tarjanSCC()	 	//O(V + E)
	{
		inSCC=new boolean [V];
		dfs_num=new int [V];
		dfs_low=new int [V];
		comp=new int [n];
		stack=new Stack<>();
		for(int i = 0; i < V; ++i)
			if(dfs_num[i] == 0)
				tarjanSCC(i);
	}
	
	static void tarjanSCC(int u)
	{
		dfs_num[u] = dfs_low[u] = ++counter;
		stack.push(u);
		
		for(int v: adjList[u])
		{
			if(dfs_num[v] == 0)
				tarjanSCC(v);
			if(!inSCC[v])
				dfs_low[u] = Math.min(dfs_low[u], dfs_low[v]);	
		}
		if(dfs_num[u] == dfs_low[u])
		{
			//SCC found
			while(true)
			{
				int v = stack.pop();
				comp[v]=SCC;
				inSCC[v] = true;
				if(v == u)
					break;
			}
			SCC++;
		}	
	}	
	
	static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {while (st == null || !st.hasMoreElements()) {
			try {st = new StringTokenizer(br.readLine());}
			catch (IOException e) {e.printStackTrace();}}
		return st.nextToken();}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() {return Double.parseDouble(next());}
		String nextLine(){String str = "";
		try {str = br.readLine();}
		catch (IOException e) {e.printStackTrace();}
		return str;}
	}
} 