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 n, m, s, cnt, SL, SL2;
vector<long long> a[6000];
bool fre[6000], bo, fre2[6000];
pair<long long, long long> CNT[6000];
void dfs(int u) {
  fre[u] = false;
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (fre[v]) dfs(v);
  }
}
void dfs2(int u) {
  SL++;
  fre2[u] = false;
  for (int i = 0; i < a[u].size(); i++) {
    int v = a[u][i];
    if (fre2[v]) dfs2(v);
  }
}
int main() {
  memset(fre, true, sizeof fre);
  memset(CNT, 0, sizeof CNT);
  cin >> n >> m >> s;
  for (int i = 1; i <= m; i++) {
    long long u, v;
    cin >> u >> v;
    a[u].push_back(v);
  }
  SL2 = 0;
  dfs(s);
  for (int i = 1; i <= n; i++) {
    if (fre[i]) {
      SL = 0;
      memset(fre2, true, sizeof fre2);
      dfs2(i);
      ++SL2;
      CNT[SL2].first = SL;
      CNT[SL2].second = i;
    }
  }
  memset(fre2, true, sizeof fre2);
  cnt = 0;
  sort(CNT + 1, CNT + SL2 + 1, greater<pair<long long, long long> >());
  for (int i = 1; i <= SL2; i++)
    if (fre2[CNT[i].second]) cnt++, dfs2(CNT[i].second);
  cout << cnt;
  return 0;
}
