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 N, M, second, cnt[(int)5000 + 7], ans;
vector<int> v[5005];
vector<pair<int, int> > vec;
bool done[(int)5000 + 7], vis[(int)5000 + 7];
void dfs(int n, bool type, int second) {
  if (type) done[n] = 1;
  vis[n] = 1;
  if (!done[n]) cnt[second]++;
  for (int i = 0; i < v[n].size(); i++) {
    int to = v[n][i];
    if (!vis[to]) dfs(to, type, second);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> N >> M >> second;
  for (int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
  }
  dfs(second, 1, 0);
  for (int i = 1; i <= N; i++) {
    if (!done[i]) {
      memset(vis, 0, sizeof vis);
      dfs(i, 0, i);
      vec.push_back(make_pair(cnt[i], i));
    }
  }
  sort(vec.begin(), vec.end());
  reverse(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    int n = vec[i].second;
    if (!done[n]) {
      memset(vis, 0, sizeof vis);
      dfs(n, 1, 0);
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
