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
sys.setrecursionlimit(6000)

n, m, s = map(int, input().split())
s -= 1
g = [[] for _ in range(n)]
for _ in range(m):
    a, b = map(int, input().split())
    g[a-1].append(b-1)

used = [False] * 5010
topo = []

def topo_sort(node):
    used[node] = True
    for c in g[node]:
        if not used[c]:
            topo_sort(c)
    topo.append(node)

def dfs(node):
    used[node] = True
    for c in g[node]:
        if not used[c]:
            dfs(c)

for i in range(n):
    if not used[i]:
        topo_sort(i)

topo.reverse()
for i in range(n):
    used[i] = False

dfs(s)
res = 0
for v in topo:
    if not used[v]:
        res += 1
        dfs(v)

print(res)