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;
long long int n, m, s, cnt[5007];
bool g[5007], vis[5007];
vector<long long int> adj[5007];
void dfs(long long int x) {
  if (!g[x]) {
    cnt[x] = 1;
  }
  vis[x] = 1;
  for (long long int i : adj[x]) {
    if (!vis[i]) {
      dfs(i);
      cnt[x] += cnt[i];
    }
  }
}
void dfs2(long long int x) {
  g[x] = 1;
  vis[x] = 1;
  for (long long int i : adj[x]) {
    if (!vis[i]) {
      dfs2(i);
    }
  }
}
bool comp(long long int x, long long int y) { return cnt[x] > cnt[y]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> s;
  for (long long int i = 0; i < m; ++i) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
  }
  queue<long long int> ben;
  ben.push(s);
  while (!ben.empty()) {
    long long int f = ben.front();
    ben.pop();
    g[f] = 1;
    for (long long int i : adj[f]) {
      if (!g[i]) {
        g[i] = 1;
        ben.push(i);
      }
    }
  }
  vector<long long int> v;
  for (long long int i = 1; i <= n; ++i) {
    if (g[i]) continue;
    v.push_back(i);
    memset(vis, 0, sizeof(vis));
    dfs(i);
  }
  long long int ans = 0;
  sort(v.begin(), v.end(), comp);
  for (long long int i = 0; i < (long long int)v.size(); ++i) {
    if (g[v[i]]) continue;
    memset(vis, 0, sizeof(vis));
    dfs2(v[i]);
    ans++;
  }
  cout << ans << '\n';
  return 0;
}
