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> A[5010], B[5010];
int n, m, c;
int vis[5010], temp[5010];
void dfs(int x) {
  vis[x] = 1;
  for (int i = 0; i < A[x].size(); ++i) {
    if (vis[A[x][i]]) continue;
    vis[A[x][i]] = 1;
    dfs(A[x][i]);
  }
}
void check(int x, int p) {
  temp[x] = 1;
  B[p].push_back(x);
  for (int i = 0; i < A[x].size(); ++i) {
    if (temp[A[x][i]]) continue;
    temp[A[x][i]] = 1;
    check(A[x][i], p);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> c;
  int x, y;
  for (int i = 0; i < m; ++i) {
    cin >> x >> y;
    A[x].push_back(y);
  }
  dfs(c);
  int remain = 0;
  priority_queue<pair<int, int> > T;
  for (int i = 1; i <= n; ++i) {
    if (vis[i]) continue;
    for (int i = 1; i <= n; ++i) temp[i] = vis[i];
    check(i, i);
  }
  for (int i = 1; i <= n; ++i) {
    if (vis[i] == 0) T.push({B[i].size(), i});
  }
  int ans = 0;
  while (T.size() > 0) {
    x = T.top().second;
    y = T.top().first;
    T.pop();
    if (vis[x]) continue;
    ans++;
    for (int i = 0; i < B[x].size(); ++i) {
      vis[B[x][i]] = 1;
    }
  }
  cout << ans << endl;
  return 0;
}
