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 deque
import heapq
import sys
input = sys.stdin.readline

def bfs1(s):
    q = deque()
    q.append(s)
    visit = [0] * (n + 1)
    visit[s] = 1
    while q:
        i = q.popleft()
        for j in G[i]:
            if not visit[j]:
                visit[j] = 1
                q.append(j)
    return visit

def bfs2(t):
    q = deque()
    q.append(t)
    visit0 = [0] * (n + 1)
    visit0[t] = 1
    c = 0
    while q:
        i = q.popleft()
        c += 1
        for j in G[i]:
            if not visit0[j] and not visit[j]:
                visit0[j] = 1
                q.append(j)
    return c

def bfs3(t):
    q = deque()
    q.append(t)
    visit[t] = 1
    while q:
        i = q.popleft()
        for j in G[i]:
            if not visit[j]:
                visit[j] = 1
                q.append(j)
    return

n, m, s = map(int, input().split())
G = [[] for _ in range(n + 1)]
for _ in range(m):
    u, v = map(int, input().split())
    G[u].append(v)
visit = bfs1(s)
h = []
for i in range(1, n + 1):
    if not visit[i]:
        c = bfs2(i)
        heapq.heappush(h, (-c, i))
ans = 0
while h:
    c, i = heapq.heappop(h)
    if not visit[i]:
        ans += 1
        bfs3(i)
print(ans)