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.
#include <bits/stdc++.h>
using namespace std;
vector<int> adj[500010], ar;
bool vis[500010];
int u, v, n, m, s, res = 0;
void dfs(int cur) {
  vis[cur] = true;
  for (auto i : adj[cur])
    if (!vis[i]) dfs(i);
  ar.push_back(cur);
}
int main() {
  scanf("%d%d%d", &n, &m, &s);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
  }
  dfs(s);
  for (int i = 1; i <= n; ++i) {
    if (!vis[i]) dfs(i);
  }
  memset(vis, false, sizeof vis);
  for (int i = n - 1; i >= 0; --i) {
    if (!vis[ar[i]]) {
      if (s != ar[i]) res++;
      dfs(ar[i]);
    }
  }
  printf("%d\n", res);
  return 0;
}
