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 = 5010;
int n, m, col[N] = {}, vis[N], cnt = 0;
vector<int> e[N];
vector<pair<int, int> > st;
void dfs1(int fder) {
  col[fder] = 1;
  int i;
  for (i = 0; i < e[fder].size(); i++) {
    if (col[e[fder][i]] == 0) dfs1(e[fder][i]);
  }
}
void dfs2(int fder) {
  vis[fder] = 1;
  cnt++;
  int i;
  for (i = 0; i < e[fder].size(); i++) {
    if (col[e[fder][i]] == 0 && vis[e[fder][i]] == 0) dfs2(e[fder][i]);
  }
}
int main() {
  int i, cap, u, v, ans = 0;
  cin >> n >> m >> cap;
  for (i = 0; i < m; i++) {
    cin >> u >> v;
    e[u - 1].push_back(v - 1);
  }
  dfs1(cap - 1);
  for (i = 0; i < n; i++) {
    if (col[i] == 0) st.push_back(make_pair(0, i));
  }
  for (i = 0; i < st.size(); i++) {
    memset(vis, 0, sizeof(vis));
    cnt = 0;
    dfs2(st[i].second);
    st[i].first = cnt;
  }
  sort(st.begin(), st.end());
  reverse(st.begin(), st.end());
  for (i = 0; i < st.size(); i++) {
    if (col[st[i].second] == 0) {
      ans += 1;
      dfs1(st[i].second);
    }
  }
  cout << ans << endl;
  return 0;
}
