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

n, m, s = map(int, input().split())

adj = [[] for _ in range(500005)]
ar = []
vis = [0] * 500005
sys.setrecursionlimit(6000)

def dfs(s):
    vis[s] = 1
    for i in range(len(adj[s])):
        if(vis[adj[s][i]] == 0):
            dfs(adj[s][i])
    ar.append(s)


for i in range(m):
    u, v = map(int, input().split())
    adj[u].append(v)

dfs(s)

for i in range(n):
    if(vis[i + 1] == 0):
        dfs(i + 1)
res = 0
vis = [0] * 500005
for i in range(n - 1, -1, -1):
    if(vis[ar[i]] == 0):
        if(s != ar[i]):
            res += 1
        dfs(ar[i])

print(res)
