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 inf = 0x7f7f7f7f;
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int n, m;
vector<int> vec[5005];
int vis[5005];
stack<int> s;
int visited[5005];
void topoSortUtil(int i) {
  visited[i] = true;
  for (int j = 0; j < vec[i].size(); j++) {
    int node = vec[i][j];
    if (!visited[node]) topoSortUtil(node);
  }
  s.push(i);
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = 0; i < vec[u].size(); i++) {
    int v = vec[u][i];
    if (vis[v] == 0) {
      dfs(v);
    }
  }
  return;
}
void display() {
  int cnt = 0;
  while (!s.empty()) {
    int val = s.top();
    s.pop();
    if (vis[val] == 0) {
      dfs(val);
      cnt++;
    }
  }
  cout << cnt << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, m, s, x, y;
  cin >> n >> m >> s;
  vector<int> ans;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    vec[x].push_back(y);
  }
  dfs(s);
  for (int i = 1; i <= n; i++) {
  }
  int cnt = 0;
  memset(visited, 0, sizeof(visited));
  for (int i = 1; i <= n; i++) {
    if (visited[i] == 0) topoSortUtil(i);
  }
  display();
  return 0;
}
