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 rev(pair<long long, long long> a, pair<long long, long long> b) {
  return a.first > b.first;
}
vector<vector<long long>> graph;
vector<bool> vis1, vis2;
void dfs_real(long long source) {
  vis1[source] = true;
  for (auto i : graph[source]) {
    if (!vis1[i]) dfs_real(i);
  }
}
long long dfs_count(long long source) {
  vis2[source] = true;
  long long cnt = 1;
  for (auto i : graph[source]) {
    if (!vis2[i]) cnt += dfs_count(i);
  }
  return cnt;
}
void solveCase() {
  long long n = 0, m = 0, s = 0;
  cin >> n >> m >> s;
  graph.resize(n);
  vis1.resize(n, false);
  while (m--) {
    long long u = 0, v = 0;
    cin >> u >> v;
    graph[u - 1].push_back(v - 1);
  }
  dfs_real(s - 1);
  vector<pair<long long, long long>> counts;
  for (long long i = 0; i < n; i++) {
    if (vis1[i]) continue;
    vis2.clear();
    vis2.resize(n, false);
    counts.push_back(make_pair(dfs_count(i), i));
  }
  sort(counts.begin(), counts.end(), rev);
  long long connected = 1;
  for (auto c : counts) {
    if (vis1[c.second]) continue;
    connected++;
    dfs_real(c.second);
  }
  cout << connected - 1 << "\n";
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solveCase();
}
