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 Contest {
	
	static ArrayList<Integer> adj[];
	static boolean[] vis;
	static boolean[] cnt;
	
	static void dfs(int node) {
		vis[node] = true;
		
		for (int v=0; v<adj[node].size(); v++) {
			int x = adj[node].get(v);
			cnt[x] = false;
			if (!vis[x]) dfs(x);
		}
	}
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		vis = new boolean[n+1];
		cnt = new boolean[n+1];
		adj = new ArrayList[n+1];
		for (int i=0; i<n+1; i++) {
			adj[i] = new ArrayList<Integer>();
		}
		for (int i=0; i<m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			adj[a].add(b);
		}
		dfs(k);
		boolean[] checked = new boolean[n+1];
		boolean[] reach = new boolean[n+1];
		for (int i=0; i<n+1; i++) {
			reach[i] = vis[i];
			checked[i] = vis[i];
		}
		for (int i=1; i<n+1; i++) {
			if (!checked[i]) {
				for (int j=0; j<n+1; j++) {
					vis[j] = reach[j];
				}
				dfs(i);
				cnt[i] = true;
				for (int j=0; j<n+1; j++) {
					if (vis[j]) checked[j] = true;
				}
			}
		}
		int s = 0;
		for (int i=0; i<n+1; i++) {
			if (cnt[i]) s++;
		}
		System.out.println(s);
	}
	
}