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>* p;
bool vis[5000];
stack<int> st;
void dfs(int src) {
  vis[src] = true;
  for (auto w : p[src]) {
    if (!vis[w]) dfs(w);
  }
}
int root(int src) {
  vis[src] = true;
  st.push(src);
  int c = 0;
  for (auto w : p[src])
    if (!vis[w]) {
      c += 1 + root(w);
    }
  return c;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, m, s, res = 0;
  cin >> n >> m >> s;
  --s;
  vector<int> g[n];
  p = g;
  for (int x, y, i = 0; i < m; ++i) {
    cin >> x >> y;
    g[--x].push_back(--y);
  }
  dfs(s);
  vector<pair<int, int>> sl;
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) {
      sl.push_back({root(i), i});
      while (st.size()) {
        vis[st.top()] = false;
        st.pop();
      }
    }
  }
  sort(sl.begin(), sl.end(), greater<pair<int, int>>());
  for (auto point : sl) {
    if (!vis[point.second]) {
      dfs(point.second);
      ++res;
    }
  }
  printf("%d\n", res);
  return false & true;
}
