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 = 5e3 + 1;
vector<pair<int, int> > E;
vector<int> g[N];
int n, m, s, cnt = 0;
set<int> S[N];
int dfn[N], low[N];
stack<int> st;
bool in[N];
void dfs(int u) {
  dfn[u] = low[u] = ++cnt;
  st.push(u);
  in[u] = 1;
  for (int i = 0; i < g[u].size(); ++i) {
    int v = g[u][i];
    if (!dfn[v]) {
      dfs(v);
      low[u] = min(low[u], low[v]);
    } else if (in[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (dfn[u] == low[u])
    while (1) {
      int v = st.top();
      st.pop();
      in[v] = 0;
      low[v] = dfn[u];
      if (v == u) break;
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin >> n >> m >> s;
  while (m--) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    E.push_back(pair<int, int>(x, y));
  }
  for (int i = 1; i <= n; ++i)
    if (!dfn[i]) dfs(i);
  for (int i = 0; i < E.size(); ++i) {
    int x = low[E[i].first];
    int y = low[E[i].second];
    if (x == y) continue;
    S[y].insert(x);
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    if (dfn[i] != low[i]) continue;
    if (!S[low[i]].size()) ans++;
  }
  ans -= (S[low[s]].size() == 0);
  cout << ans << endl;
}
