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.
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write


def in_arr():
    return map(int,raw_input().split())


def pr_num(n):
    stdout.write(str(n)+'\n')


def pr_arr(arr):
    for i in arr:
        stdout.write(str(i)+' ')
    stdout.write('\n')


range = xrange # not for python 3.0+

n,m,s=in_arr()
d=[[] for i in range(n+1)]
for i in range(m):
    u,v=in_arr()
    d[u].append(v)

good=[False]*(n+1)

q=[s]
good[s]=True
while q:
    x=q.pop()
    for i in d[x]:
        if not good[i]:
            good[i]=True
            q.append(i)
arr=[]
for i in range(1,n+1):
    if not good[i]:
        vis=list(good)
        q=[i]
        c=1
        vis[i]=1
        while q:
            x=q.pop()
            for j in d[x]:
                if not vis[j]:
                    vis[j]=True
                    q.append(j)
                    c+=1
        arr.append((c,i))
arr.sort(reverse=True)
ans=0
for _,i in arr:
    if not good[i]:
        good[i]=True
        q=[i]
        ans+=1
        while q:
            x=q.pop()
            for j in d[x]:
                if not good[j]:
                    good[j]=1
                    q.append(j)
                    
pr_num(ans)
