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;
bool pos[5001];
vector<int> susedi[5001];
vector<int> novi[5001];
queue<int> ent;
vector<pair<int, int> > iztr;
int howmany(int s) {
  while (ent.empty() == false) {
    ent.pop();
  }
  int br = 1, a = 0, b = 0, otg = 0;
  bool moipos[5001];
  for (int i = 0; i < 5001; i++) {
    moipos[i] = false;
  }
  ent.push(s);
  moipos[s] = true;
  while (br > 0) {
    a = ent.front();
    ent.pop();
    br--;
    for (int i = 0; i < susedi[a].size(); i++) {
      b = susedi[a][i];
      if (moipos[b] == false) {
        moipos[b] = true;
        if (pos[b] == false) {
          otg++;
          novi[s].push_back(b);
        }
        ent.push(b);
        br++;
      }
    }
  }
  return otg;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, s, a = 0, b = 0, br = 1, otg = 0;
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    susedi[a].push_back(b);
  }
  ent.push(s);
  pos[s] = true;
  while (br > 0) {
    a = ent.front();
    ent.pop();
    br--;
    for (int i = 0; i < susedi[a].size(); i++) {
      b = susedi[a][i];
      if (pos[b] == false) {
        pos[b] = true;
        ent.push(b);
        br++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (pos[i] == false) {
      while (ent.empty() == false) {
        ent.pop();
      }
      a = howmany(i);
      iztr.push_back({a, i});
    }
  }
  sort(iztr.begin(), iztr.end());
  for (int i = iztr.size() - 1; i >= 0; i--) {
    a = iztr[i].second;
    if (pos[a] == false) {
      pos[a] = true;
      otg++;
      for (int j = 0; j < novi[a].size(); j++) {
        b = novi[a][j];
        if (pos[b] == false) {
          pos[b] = true;
        }
      }
    }
  }
  cout << otg << "\n";
  return 0;
}
