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;
vector<int> gr[100500];
vector<int> grt[100500];
vector<int> order;
int clr[100600];
int res = 0;
int num = 0;
int compn[100500];
int good[100500];
void dfs1(int u) {
  clr[u] = 1;
  for (auto v : gr[u]) {
    if (!clr[v]) {
      dfs1(v);
    }
  }
  order.push_back(u);
}
void dfs2(int u) {
  clr[u] = 0;
  compn[u] = num;
  for (auto v : grt[u]) {
    if (clr[v]) {
      dfs2(v);
    }
    if (compn[u] != compn[v]) {
      good[compn[u]] = 1;
    }
  }
}
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  while (m--) {
    int a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    grt[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (!clr[i]) {
      dfs1(i);
    }
  }
  reverse(order.begin(), order.end());
  for (auto v : order) {
    if (clr[v]) {
      dfs2(v);
      ++num;
    }
  }
  int res = 0;
  for (int i = 0; i < num; i++) {
    res += (!good[i]) * (i != compn[s]);
  }
  cout << res << endl;
  return 0;
}
