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 = 5005;
vector<int> adj[N];
bool visited[N];
bool reach[N];
pair<int, int> has[N];
int cnt = 0;
void dfs(int s) {
  if (!visited[s]) {
    visited[s] = 1;
    reach[s] = 1;
    for (int i = 0; i < adj[s].size(); i++) {
      dfs(adj[s][i]);
    }
  }
}
void dfs2(int s, int a) {
  if (!visited[s]) {
    visited[s] = 1;
    if (reach[s] == 0) has[a].first++;
    for (int i = 0; i < adj[s].size(); i++) {
      dfs2(adj[s][i], a);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  s--;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
  }
  dfs(s);
  for (int i = 0; i < n; i++) {
    has[i].second = i;
  }
  for (int i = 0; i < n; i++) {
    if (!reach[i]) {
      memset(visited, 0, sizeof(visited));
      dfs2(i, i);
    }
  }
  sort(has, has + N);
  reverse(has, has + N);
  int ans = 0;
  cnt = 0;
  memset(visited, 0, sizeof(visited));
  for (int i = 0; i < n; i++) {
    if (!reach[has[i].second]) {
      dfs(has[i].second);
      memset(visited, 0, sizeof(visited));
      ans++;
    }
  }
  cout << ans;
  return 0;
}
