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;
bool vis[5010];
bool vis2[5010];
int cnt;
vector<vector<int>> graph(5010);
void dfs(int s) {
  vis[s] = true;
  vector<int> vizinho = graph[s];
  for (int i = 0; i < vizinho.size(); ++i) {
    if (!vis[vizinho[i]]) {
      dfs(vizinho[i]);
    }
  }
}
void dfs2(int curr) {
  cnt++;
  vis2[curr] = true;
  vector<int> vizinho = graph[curr];
  for (int i = 0; i < vizinho.size(); i++) {
    if (!vis2[vizinho[i]] and !vis[curr]) {
      dfs2(vizinho[i]);
    }
  }
}
int main() {
  int n, m, source;
  vector<pair<int, int>> arr_count;
  cin >> n >> m >> source;
  source--;
  int ans = 0;
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    --x, --y;
    graph[x].push_back(y);
  }
  dfs(source);
  for (int i = 0; i < n; i++) {
    if (!vis[i]) {
      cnt = 0;
      memset(vis2, false, sizeof(vis2));
      dfs2(i);
      arr_count.push_back(make_pair(cnt, i));
    }
  }
  sort(arr_count.begin(), arr_count.end());
  reverse(arr_count.begin(), arr_count.end());
  for (auto it : arr_count) {
    if (!vis[it.second]) {
      ans++;
      dfs(it.second);
      if (it.second == 3) {
      }
    }
  }
  cout << ans << endl;
  return 0;
}
