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.*;

public class ReachabilityFromTheCapital {
	public static void main(String[] args) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt(), m=scan.nextInt(), s=scan.nextInt()-1;
		Tarjan t=new Tarjan(n);
		for(int i=0;i<m;i++) t.add(scan.nextInt()-1,scan.nextInt()-1);
		t.go();
		int[] deg=new int[t.scc];
		for(int at=0;at<n;at++) {
			for(int nxt:t.a[at]) {
				//ah forgot to update it
				if(t.id[at]!=t.id[nxt]) {
					deg[t.id[nxt]]++;
				}
			}
		}
		int res=0;
		for(int i=0;i<t.scc;i++) {
			if(i==t.id[s]) continue;
			if(deg[i]==0) res++;
		}
		System.out.println(res);
	}
	static class Tarjan {
		int n,ct,scc;
		int[] pre,low,id;
		ArrayList<Integer>[] a;
		ArrayDeque<Integer> q;
		
		Tarjan(int n) {
			this.n=n;
			ct=0;
			scc=0;
			a=new ArrayList[n];
			for(int i=0;i<n;i++) a[i]=new ArrayList<>();
			pre=new int[n];
			low=new int[n];
			id=new int[n];
			Arrays.fill(pre,-1);
			q=new ArrayDeque<>();
		}
		
		void add(int u, int v) {
			a[u].add(v);
		}
		
		void go() {
			for(int i=0;i<n;i++) {
				if(pre[i]==-1) dfs(i);
			}
		}
		void dfs(int at) {
			pre[at]=low[at]=ct++;
			q.offer(at);
			
			for(int nxt:a[at]) {
				if(pre[nxt]==-1) dfs(nxt);
				low[at]=Math.min(low[at],low[nxt]);
			}
			
			if(low[at]==pre[at]) {
				int x=-1;
				while(x!=at) {
					x=q.pollLast();
					low[x]=n;
					id[x]=scc;
				}
				scc++;
			}
		}
	}
}