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;
int n, m, s;
vector<int> g[N];
bool used[N];
bool ok[N];
int cnt;
void dfs1(int v) {
  ok[v] = true;
  for (auto to : g[v]) {
    if (!ok[to]) {
      dfs1(to);
    }
  }
}
void dfs2(int v) {
  used[v] = true;
  ++cnt;
  for (auto to : g[v]) {
    if (!used[to] && !ok[to]) {
      dfs2(to);
    }
  }
}
int main() {
  cin >> n >> m >> s;
  --s;
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
  }
  dfs1(s);
  vector<pair<int, int>> val;
  for (int i = 0; i < n; ++i) {
    if (!ok[i]) {
      cnt = 0;
      memset(used, false, sizeof(used));
      dfs2(i);
      val.push_back({cnt, i});
    }
  }
  sort(val.rbegin(), val.rend());
  int ans = 0;
  for (auto it : val) {
    if (!ok[it.second]) {
      ++ans;
      dfs1(it.second);
    }
  }
  cout << ans << '\n';
  return 0;
}
