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 = 5005;
int n, m, s;
vector<int> g[N];
bool color[2][N];
vector<pair<int, int> > node;
int cnt;
void dfs(int idx, int from, int add = 0) {
  color[idx][from] = true;
  cnt += add;
  for (int to : g[from]) {
    if (color[idx][to]) continue;
    dfs(idx, to, add);
  }
}
void HalfDead() {
  cin >> n >> m >> s;
  for (int i = 0; i < (int)m; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
  }
  dfs(0, s);
  for (int i = 1; i <= n; ++i) {
    if (color[0][i]) continue;
    cnt = 0;
    memset(color[1], false, sizeof color[1]);
    dfs(1, i, 1);
    node.push_back({cnt, i});
  }
  sort((node).begin(), (node).end());
  reverse((node).begin(), (node).end());
  int res = 0;
  for (auto t : node) {
    int from = t.second;
    if (color[0][from]) continue;
    res += 1;
    dfs(0, from);
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  int tc = 1;
  while (tc--) HalfDead();
  return 0;
}
