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
from collections import *


class graph:
    def __init__(self):
        self.gdict = [deque() for _ in range(n + 1)]

    def addEdge(self, node1, node2, w=None):
        self.gdict[node1].append(node2)

    def dfsUtil(self, v):
        self.visit, stack = [False] * (n + 1), [v]
        self.visit[v] = True

        while (stack):
            s = stack.pop()

            for i1 in self.gdict[s]:
                if not self.visit[i1]:
                    stack.append(i1)
                    self.visit[i1] = True


rints = lambda: [int(x) for x in stdin.readline().split()]
n, m, s = rints()
g = graph()
for i in range(m):
    u, v = rints()
    g.addEdge(u, v)

g.dfsUtil(s)
bad = {x for x in range(1, n + 1) if not g.visit[x]}
cnt, ans, reach = [], 0, [[] for _ in range(n + 1)]

for x in bad:
    g.dfsUtil(x)
    tem = 0
    for i in bad:
        if g.visit[i]:
            tem += 1
            reach[x].append(i)
    cnt.append((tem,x))

cnt.sort()
while cnt:
    y,x = cnt.pop()
    dele = 0
    for i in reach[x]:
        if i in bad:
            dele = 1
            bad.discard(i)
    ans += dele
print(ans)
