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.
#Abhigyan Khaund - syl
# n = int(raw_input())
n,m,s = map(int, raw_input().split())
graph = {}
reach = {}
for i in range(1,n+1):
	graph[i] = []
	reach[i] = []
for i in range(m):
	u,v = map(int, raw_input().split())
	graph[u].append(v)

visited = [0]*(n+1)
stack = [s]

while (len(stack)!=0):
	v = stack.pop()
	if(visited[v]):
		continue
	visited[v] = 1
	for i in graph[v]:
		if(visited[i]==0):
			stack.append(i)

edges = [0]*(n+1)

for i in range(1,n+1):
	if(not visited[i]):
		# edges.append(i)
		edges[i] = 1
sot = []
for i in range(1,n+1):
	if(not edges[i]): continue
	temp_v = [0]*(n+1)
	stack = [i]
	while (len(stack)!=0):
		v = stack.pop()
		if(temp_v[v]):
			continue
		temp_v[v] = 1
		reach[i].append(v)
		for k in graph[v]:
			if(temp_v[k]==0 and edges[k]==1):
				stack.append(k)
ans = 0
order = []
for i in range(1,n+1):
	if(edges[i]==1):
		order.append([len(reach[i]),i])
order.sort(reverse=True)

for i in order:
	if(visited[i[1]]):continue
	ans+=1
	for j in reach[i[1]]:
		visited[j] = 1
print ans