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 collections import *
import sys
sys.setrecursionlimit(10**4)

n, m, s = map(int, input().split())
s -= 1
adj = [[] for _ in range(n)]
for y in range(m):
    u, v = map(int, input().split())
    adj[u-1].append(v-1)
vis = [0]*n
li = deque()
def visit(node, t=1):
    if not vis[node]:
        vis[node] = 1
        for c_node in adj[node]:
            visit(c_node)
        if t:
            li.appendleft(node)
for i in range(n):
    visit(i)
vis = [0]*n
cnt = 0
visit(s, False)
for i in list(li):
    if vis[i]:
        continue
    visit(i, False)
    cnt += 1
print(cnt)