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 double eps = 1e-13;
const int maxn = 5010;
vector<vector<int>> gr(maxn);
vector<bool> u(maxn);
vector<bool> tmp(maxn);
int cnt;
void dfs1(int v) {
  u[v] = 1;
  for (auto x : gr[v])
    if (!u[x]) dfs1(x);
}
void dfs2(int v) {
  tmp[v] = 1;
  ++cnt;
  for (auto x : gr[v]) {
    if (!tmp[x] && !u[x]) dfs2(x);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, a, b, s;
  cin >> n >> m >> s;
  --s;
  for (int i = 0; i < (int)(m); ++i) {
    cin >> a >> b;
    --a;
    --b;
    gr[a].push_back(b);
  }
  dfs1(s);
  vector<pair<int, int>> v;
  for (int i = 0; i < (int)(n); ++i) {
    if (!u[i]) {
      dfs2(i);
      tmp.assign(n, 0);
      v.push_back(make_pair(cnt, i));
      cnt = 0;
    }
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  int ans = 0;
  int k;
  for (int i = 0; i < (int)(v.size()); ++i) {
    k = v[i].second;
    if (!u[k]) {
      ++ans;
      dfs1(k);
    }
  }
  cout << ans;
  return 0;
}
