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 long long MAXN = 1e6 + 69, MOD = 1e9 + 7;
long long n, m, src, ans, tag;
long long v[MAXN], u[MAXN], mark[MAXN], comp[MAXN], deg[MAXN];
vector<long long> topol, g[MAXN], revg[MAXN];
void dfs(long long v) {
  mark[v] = 1;
  for (auto u : g[v])
    if (!mark[u]) dfs(u);
  topol.push_back(v);
}
void kosaraju(long long v) {
  mark[v] = 1, comp[v] = tag;
  for (auto u : revg[v])
    if (!mark[u]) kosaraju(u);
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> src;
  for (long long i = 1; i <= m; i++)
    cin >> v[i] >> u[i], g[v[i]].push_back(u[i]), revg[u[i]].push_back(v[i]);
  for (long long i = 1; i <= n; i++)
    if (!mark[i]) dfs(i);
  memset(mark, 0, sizeof mark);
  reverse(topol.begin(), topol.end());
  for (auto it : topol)
    if (!mark[it]) tag++, kosaraju(it);
  for (long long i = 1; i <= m; i++)
    if (comp[v[i]] != comp[u[i]]) deg[comp[u[i]]]++;
  for (long long i = 1; i <= tag; i++)
    if (i != comp[src] and deg[i] == 0) ans++;
  cout << ans << "\n";
}
