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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		Task solver = new Task();
		solver.solve(1, in, out);
		out.close();
	}

	static class Task {
		Set<Integer> canReachCities = new HashSet<>();
		List<Integer> unCities = new ArrayList<>();
		Set<Integer> tmp = new HashSet<>();
		int[] vis;

		public void solve(int testNumber, InputReader in, PrintWriter out) {
			int n = in.nextInt();
			int m = in.nextInt();
			int s = in.nextInt();
			List<Integer>[] p = new List[n + 1];

			for (int i = 1; i <= n; i++) {
				p[i] = new ArrayList<>();
			}
			for (int i = 0; i < m; i++) {
				int start = in.nextInt();
				int end = in.nextInt();
				p[start].add(end);
			}
			vis = new int[n + 1];
			DFS(p, s);
			if (canReachCities.size() == n) {
				out.println(0);
				return;
			}
			for (int i = 1; i <= n; i++) {
				if (vis[i] == 0) {
					DFS2(p, i);
				}
			}
			int count = 0;
			Set<Integer> haha = new HashSet<>();
			haha.addAll(unCities);

			vis = new int[n + 1];
			for (int i = unCities.size() - 1; i >= 0; i--) {
				if (vis[unCities.get(i)] == 0) {
					count++;
					tmp.clear();
					DFS3(p, unCities.get(i));
					haha.removeAll(tmp);
					if (haha.isEmpty()) {
						out.println(count);
						return;
					}
				}
			}
		}

		private void DFS(List<Integer>[] p, int node) {
			vis[node] = 1;
			canReachCities.add(node);
			for (Integer des : p[node]) {
				if (vis[des] == 0) {
					DFS(p, des);
				}
			}
		}

		private void DFS2(List<Integer>[] p, int node) {
			vis[node] = 1;
			for (Integer des : p[node]) {
				if (vis[des] == 0) {
					DFS2(p, des);
				}
			}
			unCities.add(node);
		}

		private void DFS3(List<Integer>[] p, int node) {
			vis[node] = 1;
			for (Integer des : p[node]) {
				if (vis[des] == 0) {
					DFS3(p, des);
				}
			}
			tmp.add(node);
		}
	}

	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

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

	}
}