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;
vector<int> v[100001];
int visited[100001], visited2[100001];
vector<pair<int, int> > v2;
void dfs(int i) {
  visited2[i] = 1;
  for (int j = 0; j < v[i].size(); ++j) {
    if (visited2[v[i][j]] == 0) dfs(v[i][j]);
  }
}
int ct;
void dfs2(int i) {
  visited[i] = 1;
  ++ct;
  for (int j = 0; j < v[i].size(); ++j) {
    if (visited[v[i][j]] == 0 && visited2[v[i][j]] == 0) dfs2(v[i][j]);
  }
}
void dfs3(int i) {
  visited2[i] = 1;
  for (int j = 0; j < v[i].size(); ++j) {
    if (visited2[v[i][j]] == 0) dfs3(v[i][j]);
  }
}
bool comp(pair<int, int> i, pair<int, int> j) { return i.second > j.second; }
int main() {
  std::ios::sync_with_stdio(false);
  int T;
  T = 1;
  while (T--) {
    int n, m, i, j, second, a, b;
    cin >> n >> m >> second;
    for (i = 0; i < m; ++i) {
      cin >> a >> b;
      v[a].push_back(b);
    }
    dfs(second);
    for (i = 1; i <= n; ++i) {
      if (visited2[i] == 0) {
        memset(visited, 0, sizeof(visited));
        ct = 0;
        dfs2(i);
        v2.push_back({i, ct});
      }
    }
    sort(v2.begin(), v2.end(), comp);
    int ans = 0;
    for (i = 0; i < v2.size(); ++i) {
      if (visited2[v2[i].first] == 1) continue;
      ++ans;
      dfs3(v2[i].first);
    }
    cout << ans << endl;
  }
  return 0;
}
