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;
void Open() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int dx[]{1, -1, 0, 0, 1, 1, -1, -1};
int dy[]{0, 0, 1, -1, 1, -1, 1, -1};
const int OO = 0x3f3f3f3f;
int n, m, s;
vector<vector<int>> g, gg;
bool p = 1;
bool pp[20002];
bool ppp[20002];
long long ans;
int en;
void dfs(int node) {
  if (pp[node]) return;
  pp[node] = 1;
  for (auto it : g[node]) dfs(it);
}
int dfs2(int node) {
  ppp[node] = 1;
  for (auto it : gg[node])
    if (ppp[it] == 0)
      return dfs2(it);
    else {
      for (auto ip : gg[it]) {
        if (ppp[ip] == 0) {
          return dfs2(ip);
        }
      }
    }
  return node;
}
void test_case() {
  cin >> n >> m >> s;
  g = vector<vector<int>>(n + 1);
  gg = vector<vector<int>>(n + 1);
  int x, y;
  for (int i = 1; i <= m; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    gg[y].push_back(x);
  }
  dfs(s);
  for (int i = 1; i <= n; i++) {
    if (!pp[i]) {
      int x = dfs2(i);
      dfs(x);
      ans++;
    }
  }
  cout << ans << "\n";
}
int main() {
  Open();
  int t = 1;
  while (t--) {
    test_case();
  }
}
