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 sys
sys.setrecursionlimit(1000000)
 

n, m, s = [int(x) for x in input().split()]
adj = [[] for u in range(n + 1)]
vis = [False for u in range(n + 1)]
for i in range(m):
    u, v = [int(x) for x in input().split()]
    adj[u].append(v)

def tarjan():
    low = [-1 for u in range(n + 1)]
    disc = [-1 for v in range(n + 1)]
    in_stack = [False for v in range(n + 1)]
    st = deque()
    comp = []

    def dfs(u):
        dfs.dfs_time = dfs.dfs_time + 1
        low[u] = dfs.dfs_time
        disc[u] = dfs.dfs_time
        in_stack[u] = True
        st.append(u)

        for v in adj[u]:
            if disc[v] == -1:
                dfs(v)
                low[u] = min(low[u], low[v])
            elif in_stack[v]:
                low[u] = min(low[u], disc[v])

        if low[u] == disc[u]:
            comp.append([])
            while u != st[-1]:
                x = st.pop()
                in_stack[x] = False
                comp[-1].append(x)
            x = st.pop()
            in_stack[x] = False
            comp[-1].append(x)

    dfs.dfs_time = 0
    for i in range(1, n + 1):
        if disc[i] == -1:
            dfs(i)

    return comp[::-1]

def visit(adj, vis, src):
    q = deque()
    q.append(src)
    vis[src] = True
    while len(q) > 0: 
        u = q.popleft()
        for v in adj[u]:
            if vis[v] == False:
                vis[v] = True
                q.append(v)
    
ans = 0
visit(adj, vis, s)
tarOrd = tarjan()
for lis in tarOrd:
    x = lis[0]
    if vis[lis[0]] == False:
        visit(adj, vis, x)
        ans += 1

print(ans)



