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 sys
sys.setrecursionlimit(100001)
from collections import *
d=deque()
def dfs2(start):
    vis[start]=1
    for child in adj[start]:
        if vis[child]==0:
            dfs2(child)

def dfs(start):
    vis[start]=1
    for child in adj[start]:
        if vis[child]==0:
            dfs(child)
    d.appendleft(start)

n,m,k=map(int,input().split())

adj=[]
for i in range(n):
    adj.append([])

for i in range(m):
    a,b=map(int,input().split())
    adj[a-1].append(b-1)

vis=[0]*n
for i in range(n):
    if vis[i]==0:
        dfs(i)
vis=[0]*n

dfs2(k-1)
cnt=0
for i in d:
    if vis[i]==0:
        dfs2(i)
        cnt+=1
print(cnt)