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;
vector<int> g[5000];
vector<int> r[5000];
vector<int8_t> vis(5000);
int n, m, s;
void dfs(int src) {
  vis[src] = true;
  for (auto v : g[src]) {
    if (!vis[v]) {
      dfs(v);
    }
  }
}
int root(int src, vector<bool> &df) {
  df[src] = true;
  int c = 0;
  for (int v : g[src]) {
    if (!df[v] && !vis[v]) {
      c += root(v, df) + 1;
    }
  }
  return c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> s;
  --s;
  for (int x, y, i = 0; i < m; ++i) {
    cin >> x >> y;
    g[--x].push_back(--y);
    r[y].push_back(x);
  }
  dfs(s);
  int ans = 0;
  vector<pair<int, int>> sl;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      vector<bool> df(n);
      int a = root(i, df);
      sl.push_back({-a, i});
    }
  }
  sort(sl.begin(), sl.end());
  for (auto c : sl) {
    if (!vis[c.second]) {
      g[s].push_back(c.second);
      dfs(s);
      ++ans;
    }
  }
  printf("%d\n", ans);
  return false & true;
}
