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 long long MAX = 1e5;
vector<pair<long long, long long> > edges;
vector<long long> topo, vec[MAX], rev[MAX];
long long N, M, S, X, Y, cur, res, deg[MAX], mark[MAX];
void dfs(long long node) {
  mark[node] = 1;
  for (auto A : vec[node])
    if (!mark[A]) dfs(A);
  topo.push_back(node);
  return;
}
void dfs(long long node, long long id) {
  mark[node] = id;
  for (auto A : rev[node])
    if (!mark[A]) dfs(A, id);
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> M >> S;
  for (long long A = 1; A <= M; A++) {
    cin >> X >> Y;
    vec[X].push_back(Y);
    rev[Y].push_back(X);
    edges.push_back(make_pair(X, Y));
  }
  for (long long A = 1; A <= N; A++)
    if (!mark[A]) dfs(A);
  for (long long A = 1; A <= N; A++) mark[A] = 0;
  while (!topo.empty()) {
    cur = topo.back();
    topo.pop_back();
    if (mark[cur]) continue;
    dfs(cur, cur);
  }
  for (auto A : edges) {
    if (mark[A.first] != mark[A.second]) {
      deg[mark[A.second]]++;
    }
  }
  for (long long A = 1; A <= N; A++)
    if (mark[A] == A or mark[A] == 0) res += (!deg[A]);
  res -= (!deg[mark[S]]);
  cout << res;
  return 0;
}
