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;
const int mod = 1e9 + 7;
vector<long long int> adj[5000 + 1];
long long int vis[5000 + 1];
void dfs(long long int v, long long int par) {
  vis[v] = par;
  for (auto i : adj[v]) {
    if (vis[i] != par) dfs(i, par);
  }
}
void solve() {
  long long int N, M, S;
  cin >> N >> M >> S;
  for (long long int i = 0; i < M; ++i) {
    long long int u, v;
    cin >> u >> v;
    if (v != S) adj[u].push_back(v);
  }
  for (long long int i = 1; i <= N; ++i)
    if (vis[i] == 0) dfs(i, i);
  set<long long int> ans;
  for (long long int i = 1; i <= N; ++i) ans.insert(vis[i]);
  cout << ans.size() - 1 << '\n';
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  while (T--) {
    solve();
  }
  exit(0);
}
