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 S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
  return pair<S, T>(s.first + t.first, s.second + t.second);
}
template <class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) {
  return pair<S, T>(s.first - t.first, s.second - t.second);
}
template <class S, class T>
ostream &operator<<(ostream &os, pair<S, T> p) {
  os << "(" << p.first << ", " << p.second << ")";
  return os;
}
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 (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const long long inf = 1000000001;
const long long INF = 2e18;
const long long MOD = 1000000007;
const double pi = 3.14159265358979323846;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int dx2[8] = {1, 1, 0, -1, -1, -1, 0, 1}, dy2[8] = {0, 1, 1, 1, 0, -1, -1, -1};
vector<vector<int> > child(5010), parent(5010);
vector<int> color(5010, -1);
int n, m, s;
vector<int> roots, roots2;
void dfs1(int now, int col) {
  color[now] = col;
  for (long long i = 0; i < (long long)(child[now].size()); i++) {
    if (color[child[now][i]] < 0) {
      dfs1(child[now][i], col);
    }
  }
}
void dfs2(int now) {
  color[now] = -2;
  if (parent[now].size() == 0) {
    roots.push_back(now);
    return;
  }
  bool flag = true;
  for (long long i = 0; i < (long long)(parent[now].size()); i++) {
    if (color[parent[now][i]] == -1) {
      dfs2(parent[now][i]);
      flag = false;
    }
  }
  if (flag) {
    roots2.push_back(now);
  }
  return;
}
int main() {
  cin >> n >> m >> s;
  s--;
  for (long long i = 0; i < (long long)(m); i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    child[u].push_back(v);
    parent[v].push_back(u);
  }
  dfs1(s, s);
  int ans = 0;
  for (long long i = 0; i < (long long)(n); i++) {
    if (color[i] >= 0) {
      continue;
    }
    roots.clear();
    dfs2(i);
    ans += roots.size();
    for (long long j = 0; j < (long long)(roots.size()); j++) {
      dfs1(roots[j], roots[j]);
    }
  }
  for (long long i = 0; i < (long long)(roots2.size()); i++) {
    if (color[roots2[i]] < 0) {
      ans++;
      dfs1(roots2[i], roots2[i]);
    }
  }
  cout << ans << endl;
}
