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>
bool chmax(T& a, const T& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
const long long int MOD = 1e9 + 7, INF = 1e18;
long long int dx[8] = {0, -1, 1, 0, 1, -1, 1, -1},
              dy[8] = {1, 0, 0, -1, -1, -1, 1, 1};
struct edge {
  long long int cost;
  long long int u, v;
};
template <typename G>
struct StronglyConnectedComponents {
  const G& g;
  vector<vector<long long int>> gg, rg;
  vector<int> comp, order, used;
  StronglyConnectedComponents(G& g)
      : g(g), gg(g.size()), rg(g.size()), comp(g.size(), -1), used(g.size()) {
    for (int i = 0; i < g.size(); i++) {
      for (auto e : g[i]) {
        gg[i].emplace_back((int)e);
        rg[(int)e].emplace_back(i);
      }
    }
  }
  int operator[](int k) { return comp[k]; }
  void dfs(int idx) {
    if (used[idx]) return;
    used[idx] = true;
    for (int to : gg[idx]) dfs(to);
    order.push_back(idx);
  }
  void rdfs(int idx, int cnt) {
    if (comp[idx] != -1) return;
    comp[idx] = cnt;
    for (int to : rg[idx]) rdfs(to, cnt);
  }
  void build(vector<vector<long long int>>& t) {
    for (int i = 0; i < gg.size(); i++) dfs(i);
    reverse(begin(order), end(order));
    int ptr = 0;
    for (int i : order)
      if (comp[i] == -1) rdfs(i, ptr), ptr++;
    t.resize(ptr);
    for (int i = 0; i < g.size(); i++) {
      for (auto& to : g[i]) {
        int x = comp[i], y = comp[to];
        if (x == y) continue;
        t[x].push_back(y);
      }
    }
  }
};
long long int N, M, s, u, v;
int main() {
  cin >> N >> M >> s;
  s--;
  vector<vector<long long int>> to(N, vector<long long int>()), buff;
  for (long long int i = (0), i_end_ = (M); i < i_end_; i++) {
    cin >> u >> v;
    u--;
    v--;
    to[v].push_back(u);
  }
  long long int ans = 0;
  StronglyConnectedComponents<vector<vector<long long int>>> g(to);
  g.build(buff);
  for (long long int i = (0), i_end_ = (((long long int)(buff).size()));
       i < i_end_; i++) {
    if (i == g[s]) continue;
    if (((long long int)(buff[i]).size()) == 0) ans++;
  }
  cout << ans << '\n';
}
