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 par[5005];
int check[5005];
void dfs(vector<int> vt[], int s, int src, bool vis[]) {
  vis[s] = 1;
  par[s] = src;
  for (int i = 0; i < vt[s].size(); i++) {
    if (!vis[vt[s][i]]) dfs(vt, vt[s][i], src, vis);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int m, n, s, i, j, k, u, v;
  cin >> n >> m >> s;
  vector<int> vt[n + 1];
  while (m--) {
    cin >> u >> v;
    vt[u].push_back(v);
  }
  bool gvis[5005];
  bool vis[5005];
  memset(gvis, 0, sizeof(gvis));
  dfs(vt, s, s, gvis);
  int cou = 0;
  for (i = 1; i <= n; i++) {
    if (i != s && par[i] == 0) {
      memset(vis, 0, sizeof(vis));
      dfs(vt, i, i, vis);
    }
  }
  for (i = 1; i <= n; i++) {
    if (par[i] == 0) cou++;
  }
  for (i = 1; i <= n; i++) {
    if (par[i] != s && !check[par[i]]) {
      cou++;
      check[par[i]] = 1;
    }
  }
  cout << cou << "\n";
  return 0;
}
