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 __future__ import print_function
from queue import Queue
from random import shuffle
import sys
import math
import os.path


sys.setrecursionlimit(100000)

# LOG

def log(*args, **kwargs):
    print(*args, file=sys.stderr, **kwargs)


# INPUT

def ni():
    return map(int, input().split())


def nio(offset):
    return map(lambda x: int(x) + offset, input().split())


def nia():
    return list(map(int, input().split()))


# CONVERT
def toString(aList, sep=" "):
    return sep.join(str(x) for x in aList)


def toMapInvertIndex(aList):
    return {k: v for v, k in enumerate(aList)}


# SORT
def sortId(arr):
    return sorted(range(len(arr)), key=lambda k: arr[k])


# MAIN

n,m,s = ni()

s -= 1

adj = [[] for _ in range(n)]

for i in range(m):
    u,v = nio(-1)
    if (v != s):
        adj[u].append(v)

stack = []

visited= [False]*n

def dfs(x):
    global visited
    global stack
    visited[x] = True
    for y in adj[x]:
        if not visited[y]:
            dfs(y)

    stack.append(x)


for i in range(n):
    if not visited[i]:
        dfs(i)

# log(adj)
# log(visited)
# log(stack)

count = -1

def loang(x):
    global visited
    visited[x] = False
    for y in adj[x]:
        if visited[y]:
            loang(y)


for x in stack[::-1]:
    if visited[x]:
        count += 1
        loang(x)

print(count)