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;
const int N = 5e3 + 100;
vector<int> g[N];
bool mark[N], reach[N];
set<int> st;
void dfs(int v) {
  mark[v] = reach[v] = true;
  for (auto u : g[v]) {
    if (!mark[u]) dfs(u);
    if (st.find(u) != st.end()) st.erase(u);
  }
  return;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
  }
  dfs(s);
  for (int i = 1; i <= n; i++)
    if (!reach[i]) {
      dfs(i);
      st.insert(i);
      memset(mark, 0, sizeof mark);
    }
  return cout << st.size() << "\n", 0;
}
