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

def dfs_visit(s, Adj, visited):
	for v in Adj[s]:
		if not visited[v]:
			visited[v] = s
			dfs_visit(v, Adj, visited)
	can_reach.append(s)


n, m, cap = map(int, raw_input().split())
visited = [False  for zz in xrange(n+1)]
Adj = [[] for num in xrange(n+1)]
extra = 0
can_reach = []
visited[cap] = True
sys.setrecursionlimit(1000000)


for x in xrange(m):
	u, v = map(int, raw_input().split())
	Adj[u].append(v)


dfs_visit(cap, Adj, visited)

for x in xrange(1, len(visited)):
	if not visited[x]:
		visited[x] = True
		dfs_visit(x, Adj, visited)

visited = [False  for zz in xrange(n+1)]
visited[cap] = True
dfs_visit(cap, Adj, visited)


for x in can_reach[::-1]:
	if not visited[x]:
		extra += 1
		dfs_visit(x, Adj, visited)

print extra








