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 maxn = 1e4;
bool mark[maxn];
int deg[maxn], comp[maxn], compcnt, n, m, s;
vector<int> adj[maxn], jda[maxn];
stack<int> stc;
void dfs(int u) {
  mark[u] = true;
  for (auto v : adj[u])
    if (!mark[v]) dfs(v);
  stc.push(u);
}
void sfd(int u) {
  mark[u] = true;
  comp[u] = compcnt;
  for (auto v : jda[u])
    if (!mark[v])
      sfd(v);
    else if (comp[v] != compcnt)
      deg[compcnt]++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> s;
  s--;
  for (int i = 0; i < m; i++) {
    int fi, se;
    cin >> fi >> se;
    fi--, se--;
    adj[fi].push_back(se);
    jda[se].push_back(fi);
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) dfs(i);
  memset(mark, false, sizeof mark);
  while (((int)(stc).size())) {
    int u = stc.top();
    stc.pop();
    if (mark[u]) continue;
    sfd(u);
    compcnt++;
  }
  int ans = 0;
  for (int i = 0; i < compcnt; i++)
    if (deg[i] == 0 and i != comp[s]) ans++;
  cout << ans << endl;
  return 0;
}
