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 = 5001;
bool yes[N];
vector<int> adj[N];
void dfs(int u) {
  if (yes[u]) return;
  yes[u] = 1;
  for (auto el : adj[u]) dfs(el);
}
bool vis[N];
priority_queue<pair<int, int>> ss;
int cnt(int u) {
  if (vis[u]) return 0;
  vis[u] = 1;
  int res = 0;
  for (auto el : adj[u])
    if (!yes[el]) res += cnt(el);
  return res + 1;
}
void yess(int u) {
  if (yes[u]) return;
  yes[u] = 1;
  for (auto el : adj[u]) yess(el);
}
int main() {
  int n, m, s;
  scanf("%d %d %d", &n, &m, &s);
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &u, &v);
    u--, v--;
    adj[u].push_back(v);
  }
  s--;
  dfs(s);
  for (int i = 0; i < n; i++) {
    if (!yes[i]) {
      memset(vis, 0, sizeof(vis));
      ss.push({cnt(i), i});
    }
  }
  int res = 0;
  while (!ss.empty()) {
    pair<int, int> pr = ss.top();
    ss.pop();
    if (yes[pr.second]) continue;
    res++;
    yess(pr.second);
  }
  cout << res;
}
