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 A
{	
		
		
	static int []SCCof,dfsNum,dfsLow;
	static ArrayList<Integer> []adj,DAG;
	static Stack<Integer> stack;
	static boolean []inSCC;
	static int counter,SCC;
	static boolean []visited;
	public static void solve(int n)
	{
		dfsNum=new int [n];
		dfsLow=new int [n];
		inSCC=new boolean[n];
		SCCof=new int [n];
		stack=new Stack();
		for(int i=0;i<n;i++)
			if(dfsNum[i]==0)
				SCC(i);
	}
	public static void SCC(int u)
	{
		dfsNum[u]=dfsLow[u]=++counter;
		
		stack.push(u);
		for(int v:adj[u])
		{
			if(dfsNum[v]==0)
				SCC(v);
			if(!inSCC[v])
				dfsLow[u]=Math.min(dfsLow[u], dfsLow[v]);
		}
		if(dfsNum[u]==dfsLow[u])
		{
			
			while(true)
			{
				int v=stack.pop();
				
				SCCof[v]=SCC;
				inSCC[v]=true;
				if(u==v)
					break;
			}
			
			SCC++;
		
		}
	}
		public static void dfs(int u)
		{
			visited[u]=true;
			for(int v:DAG[u])
				if(!visited[v])
					dfs(v);
		}
		public static void main(String[] args) throws IOException 
		{
				Scanner sc=new Scanner(System.in);
				PrintWriter pw=new PrintWriter(System.out);
				int n=sc.nextInt(),m=sc.nextInt(),s=sc.nextInt()-1;
				adj=new ArrayList[n];
				for(int i=0;i<n;i++)
					adj[i]=new ArrayList();
				while(m-->0)
					adj[sc.nextInt()-1].add(sc.nextInt()-1);
				solve(n);
				DAG=new ArrayList[SCC];
				for(int i=0;i<SCC;i++)
					DAG[i]=new ArrayList();
				int []deg=new int [SCC];
				for(int i=0;i<n;i++)
				{
					int u=SCCof[i];
					for(int x:adj[i])
					{
						int v=SCCof[x];
						if(u!=v)
							deg[v]++;
					}
				}
				visited=new boolean[SCC];
				
				/*dfs(SCCof[s]);*/
				int ans=0;
				for(int i=0;i<SCC;i++)
					if(deg[i]==0 && i!=SCCof[s])
						ans++;
				pw.println(ans);
				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 Scanner(String s) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(s));
		}
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		
		
		
}