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;
using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
const int N = 5001;
const int MOD = 1e9 + 7;
const ll INF = 1e18 + 1;
int n, m, cap;
vector<int> adj[N], adjt[N];
vector<int> order;
vector<int> visited(N, 0), good(N, 0);
vector<int> comp(N);
int cmp = 0;
void dfs1(int i) {
  if (visited[i]) return;
  visited[i] = 1;
  for (auto j : adj[i])
    if (!visited[j]) dfs1(j);
  order.emplace_back(i);
}
void dfs2(int i) {
  comp[i] = cmp;
  if (visited[i]) return;
  visited[i] = 1;
  for (auto j : adjt[i])
    if (!visited[j]) dfs2(j);
}
void kosaraju() {
  for (int i = 1; i <= n; i++)
    if (!visited[i]) dfs1(i);
  reverse(order.begin(), order.end());
  order.insert(order.begin(), 0);
  visited.assign(N, false);
  for (int i = 1; i <= n; i++)
    if (!visited[order[i]]) {
      dfs2(order[i]);
      cmp++;
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> cap;
  for (int i = 0; i < m; i++) {
    int s, d;
    cin >> s >> d;
    adj[s].emplace_back(d);
    adjt[d].emplace_back(s);
  }
  kosaraju();
  set<int> ans;
  for (int i = 0; i < cmp; i++) ans.insert(i);
  ans.erase(comp[cap]);
  for (int u = 1; u <= n; u++)
    for (auto v : adj[u])
      if (comp[u] != comp[v]) ans.erase(comp[v]);
  cout << ans.size();
  return 0;
}
