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.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
import java.io.*;

public class Test5{

    static ArrayList<Integer>[] l;
    static boolean[] bol, bol2;
    static int[] remv, remk;
    static int g=0;
    static ArrayList<Integer> vert = new ArrayList<>();

    public static void main(String[] z) throws Exception{
        Scanner s = new Scanner(System.in);
        int a = s.nextInt(), b=s.nextInt(), stv=s.nextInt()-1, o=0,x,y, t=0;
        l = new ArrayList[a];
        bol = new boolean[a];
        bol2 = new boolean[a];
        for(int q=0; q<a; q++) l[q] = new ArrayList<>();
        for(int q=0; q<b; q++){
            x = s.nextInt();
            y = s.nextInt();
            l[x-1].add(y-1);
        }
        dfs(stv);
        for(int q=0; q<a; q++) if(!bol[q]) vert.add(q);
        remv = new int[vert.size()];
        remk = new int[vert.size()];
        for(int u : vert){
            g=0;
            bol2=Arrays.copyOf(bol, a);
            dfs2(u);
            remv[t] = u;
            remk[t] = g;
            t++;
        }
        PyraSort.sort(remk);
        for(int q=vert.size()-1; q>=0; q--){
            if(!bol[remv[q]]){
                dfs(remv[q]);
                o++;
            }
        }
        System.out.print(o);
    }

    static void dfs(int v){
        bol[v] = true;
        bol2[v] = true;
        for(int h : l[v]) if(!bol[h]) dfs(h);
    }

    static void dfs2(int v){
        bol2[v] = true;
        g++;
        for(int h : l[v]) if(!bol2[h]) dfs2(h);
    }

    static class PyraSort {

        private static int heapSize;

        public static void sort(int[] a) {
            buildHeap(a);
            while (heapSize > 1) {
                swap(a, 0, heapSize - 1);
                heapSize--;
                heapify(a, 0);
            }
        }

        private static void buildHeap(int[] a) {
            heapSize = a.length;
            for (int i = a.length / 2; i >= 0; i--) {
                heapify(a, i);
            }
        }

        private static void heapify(int[] a, int i) {
            int l = 2 * i + 2;
            int r = 2 * i + 1;
            int largest = i;
            if (l < heapSize && a[i] < a[l]) {
                largest = l;
            }
            if (r < heapSize && a[largest] < a[r]) {
                largest = r;
            }
            if (i != largest) {
                swap(a, i, largest);
                heapify(a, largest);
            }
        }

        private static void swap(int[] a, int i, int j) {
            a[i] ^= a[j] ^= a[i];
            a[j] ^= a[i];
            remv[i] ^= remv[j] ^= remv[i];
            remv[j] ^= remv[i];
        }
    }
}
