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;
template <class T>
T sqr(T x) {
  return x * x;
}
const int maxn = 6001;
vector<int> g[maxn];
int n, m, s;
int x, y;
vector<pair<int, int> > lst;
int vis[maxn], vis_cur[maxn], mark[maxn];
void mark_dfs(int x) {
  mark[x] = 1;
  for (auto &v : g[x]) {
    if (mark[v]) continue;
    mark_dfs(v);
  }
}
int cnt = 0;
void dfs(int x) {
  vis[x] = 1;
  cnt++;
  for (auto &v : g[x]) {
    if (mark[v] | vis[v]) continue;
    dfs(v);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &s);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
  }
  mark_dfs(s);
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    for (int j = 1; j <= n; j++) vis[j] = 0;
    cnt = 0;
    dfs(i);
    lst.push_back(pair<int, int>(cnt, i));
  }
  sort(lst.begin(), lst.end());
  reverse(lst.begin(), lst.end());
  int ans = 0;
  for (int i = 0; i < lst.size(); i++) {
    int x = lst[i].second;
    if (mark[x]) continue;
    mark_dfs(x);
    ans++;
  }
  cout << ans << endl;
}
