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 N = (int)3e5 + 5;
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
const int unvisit = 0;
const int visit = 1;
int n, m, s, cnt;
int a[N], vs[N], used[N];
vector<int> graph[N];
void init() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
void dfs1(int u) {
  vs[u] = 1;
  for (__typeof(graph[u].begin()) it = graph[u].begin(); it != graph[u].end();
       it++) {
    int v = *it;
    if (!vs[v]) dfs1(v);
  }
}
void dfs2(int u) {
  used[u] = 1;
  cnt++;
  for (__typeof(graph[u].begin()) it = graph[u].begin(); it != graph[u].end();
       it++) {
    int v = *it;
    if (!used[v] && !vs[v]) {
      dfs2(v);
    }
  }
}
void solve() {
  cin >> n >> m >> s;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
  }
  dfs1(s);
  vector<pair<int, int> > val;
  for (int i = 1; i <= n; i++) {
    if (!vs[i]) {
      cnt = 0;
      memset(used, 0, sizeof(used));
      dfs2(i);
      val.push_back({cnt, i});
    }
  }
  sort(val.begin(), val.end());
  reverse(val.begin(), val.end());
  int ans = 0;
  for (__typeof(val.begin()) it = val.begin(); it != val.end(); it++) {
    pair<int, int> v = *it;
    if (!vs[v.second]) {
      ans++;
      dfs1(v.second);
    }
  }
  cout << ans << '\n';
}
int main() {
  init();
  solve();
  return 0;
}
