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 double PI = acos(-1.0);
const double EPS = -1e-2;
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
const long long mod = 1e9 + 7;
const int Mx = INT_MAX;
const int Mn = INT_MIN;
const long long MX = LLONG_MAX;
const long long MN = LLONG_MIN;
const int N = 2e5 + 2;
void Open() {}
int cnt = 0;
vector<vector<int>> adj;
bitset<N> vis, vis2;
void dfs(int node) {
  vis[node] = 1;
  for (int next : adj[node]) {
    if (!vis[next]) dfs(next);
  }
}
void dfs2(int node) {
  vis2[node] = 1;
  cnt++;
  for (int next : adj[node]) {
    if (!vis2[next]) dfs2(next);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Open();
  int n, m, s, u, v;
  cin >> n >> m >> s;
  adj.resize(n + 1);
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
  }
  dfs(s);
  vector<pair<int, int>> city;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      vis2.reset();
      cnt = 0;
      dfs2(i);
      city.push_back({cnt, i});
    }
  }
  int ans = 0;
  sort(((city).rbegin()), ((city).rend()));
  for (auto i : city) {
    if (!vis[i.second]) {
      ans++;
      adj[s].push_back(i.second);
      dfs(i.second);
    }
  }
  cout << ans;
}
