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;
int TM;
int n, m, s, cl;
vector<int> G[5003], Q[5003];
int vis[5003];
bool ok[5003];
bool fcc, flc;
void chk(int u) {
  if (ok[u]) return;
  ok[u] = 1;
  for (int i = 0; i < Q[u].size(); i++) chk(Q[u][i]);
  return;
}
int dfs(int u) {
  if (u == s) return 0;
  if (vis[u]) {
    if (vis[u] == cl)
      fcc = 1;
    else
      flc = 1;
    return 0;
  }
  vis[u] = cl;
  if (G[u].size() == 0) return 1;
  int res = 0;
  for (int i = 0; i < G[u].size(); i++) res += dfs(G[u][i]);
  return res;
}
int main() {
  int u, v;
  cin >> n >> m >> s;
  while (m--) {
    cin >> u >> v;
    Q[u].push_back(v);
    G[v].push_back(u);
  }
  chk(s);
  int ans = 0;
  for (int u = 1; u <= n; u++) {
    if (ok[u]) continue;
    if (vis[u]) continue;
    cl++;
    fcc = 0;
    flc = 0;
    int cur = dfs(u);
    ans += cur;
    if (cur == 0 && fcc && !flc) ans++;
  }
  cout << ans;
}
