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 P = pair<int, int>;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("Ofast")
constexpr ll INF = 1e18;
constexpr int inf = 1e9;
constexpr double INFD = 1e100;
constexpr ll mod = 1000000007;
constexpr ll mod2 = 998244353;
const double PI = 3.1415926535897932384626433832795028841971;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int N, M, s;
vector<bool> ok(5010, false);
vector<vector<int>> G(5010);
void dfs(int v) {
  ok[v] = true;
  for (int nv : G[v]) {
    if (ok[nv]) continue;
    dfs(nv);
  }
}
vector<bool> used(5010, false);
int cnt = 0;
void dfs2(int v) {
  used[v] = true;
  cnt++;
  for (int nv : G[v]) {
    if (!used[nv] and !ok[nv]) {
      dfs2(nv);
    }
  }
}
int main() {
  cin >> N >> M >> s;
  s--;
  for (int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    G[x].push_back(y);
  }
  dfs(s);
  vector<P> rem;
  for (int i = 0; i < N; i++) {
    if (ok[i]) continue;
    used.assign(N, false);
    cnt = 0;
    dfs2(i);
    rem.emplace_back(cnt, i);
  }
  sort(rem.begin(), rem.end(), greater<P>());
  int ans = 0;
  for (int i = 0; i < rem.size(); i++) {
    if (ok[rem[i].second]) continue;
    ans++;
    dfs(rem[i].second);
  }
  cout << ans << '\n';
  ;
  return 0;
}
