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;
void matar(string NotaSuicida) {
  cout << NotaSuicida;
  exit(0);
}
const long long N = 5e3 + 15, mod = 1e9 + 7, lim = 1e18 + 5ll;
int n, m, s, a, b, c;
vector<int> g[N];
vector<pair<int, int> > pq;
bool vs[N];
int h[N];
bool kk[N];
stack<int> st;
void DFS(int u) {
  st.push(u);
  vs[u] = true;
  c++;
  for (int v : g[u])
    if (!vs[v]) DFS(v);
  return;
}
void gg(int u) {
  vs[u] = true;
  for (int v : g[u])
    if (!vs[v]) gg(v);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    g[a].push_back(b);
  }
  gg(s);
  for (int i = 1; i <= n; i++)
    if (!vs[i]) {
      c = 0;
      DFS(i);
      h[i] = c;
      while (!st.empty()) {
        vs[st.top()] = false;
        st.pop();
      }
      pq.push_back({h[i], i});
    }
  sort(pq.begin(), pq.end());
  reverse(pq.begin(), pq.end());
  int k = 0;
  for (int i = 0; i < ((int)pq.size()); i++) {
    if (vs[pq[i].second]) continue;
    k++;
    DFS(pq[i].second);
  }
  cout << k;
  return 0;
}
