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 INF = 0x3f3f3f3f;
const int maxn = 5005;
string str;
int l;
vector<pair<int, int> > v;
vector<int> e[maxn];
int n, m, cap;
int good[maxn];
int subflg = 0;
int ans = 0;
int du[maxn];
void dfs(int i) {
  if (good[i]) {
    return;
  }
  good[i] = 1;
  for (int j = 0; j < e[i].size(); j++) {
    dfs(e[i][j]);
  }
}
int cnt[maxn] = {0};
int now;
int vis[maxn] = {0};
int bad_dfs(int i) {
  if (vis[i]) return 0;
  vis[i] = 1;
  if (!good[i]) cnt[now]++;
  for (int j = 0; j < e[i].size(); j++) {
    bad_dfs(e[i][j]);
  }
  return cnt[now];
}
int main() {
  memset(good, 0, sizeof(good));
  cin >> n >> m >> cap;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
  }
  dfs(cap);
  for (int i = 1; i <= n; i++) {
    if (good[i]) continue;
    memset(vis, 0, sizeof(vis));
    now = i;
    bad_dfs(i);
  }
  int iter[maxn];
  for (int i = 1; i <= n; i++) iter[i] = i;
  sort(iter + 1, iter + n + 1, [](int x, int y) { return cnt[x] > cnt[y]; });
  for (int i = 1; i <= n; i++) {
    int now = iter[i];
    subflg = 0;
    if (!good[now]) {
      dfs(now);
      ans++;
    }
  }
  cout << ans << endl;
}
