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 = 5e3 + 1;
vector<int> ok[N];
vector<pair<int, int> > go;
int vis[N], vis2[N], path[N];
void dfs(int node, int start, int t) {
  if (!t) vis[node] = 1;
  if (t) vis2[node] = 1;
  for (int i = 0; i < ok[node].size(); i++) {
    if (!vis[ok[node][i]] && !t) dfs(ok[node][i], start, t);
    if (!vis2[ok[node][i]] && t && !vis[ok[node][i]]) {
      path[start]++;
      dfs(ok[node][i], start, t);
    }
  }
}
int main() {
  int n, m, x;
  scanf("%d%d%d", &n, &m, &x);
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    ok[u].push_back(v);
  }
  dfs(x, 0, 0);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      for (int j = 1; j <= n; j++) vis2[j] = 0;
      dfs(i, i, 1);
      go.push_back(make_pair(-path[i], i));
    }
  }
  sort(go.begin(), go.end());
  for (int i = 0; i < go.size(); i++)
    if (!vis[go[i].second]) {
      ans++;
      dfs(go[i].second, 0, 0);
    }
  printf("%d\n", ans);
  return 0;
}
