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;
int n, m, s, d[(int)(1e5 + 4)], sol, id, ctc[(int)(1e5 + 4)];
bool used[(int)(1e5 + 4)];
vector<int> g[(int)(1e5 + 4)], tg[(int)(1e5 + 4)], post;
void build(int node) {
  used[node] = 1;
  for (auto it : g[node])
    if (!used[it]) build(it);
  post.push_back(node);
}
void dfs(int node) {
  used[node] = 1;
  ctc[node] = id;
  for (auto it : tg[node])
    if (!used[it]) dfs(it);
}
int main() {
  cin >> n >> m >> s;
  for (int i = 0, x, y; i < m; ++i) {
    cin >> x >> y;
    g[x].push_back(y);
    tg[y].push_back(x);
  }
  for (int i = 1; i <= n; ++i)
    if (!used[i]) build(i);
  memset(used, 0, sizeof used);
  reverse(post.begin(), post.end());
  for (auto it : post)
    if (!used[it]) {
      ++id;
      dfs(it);
    }
  for (int i = 1; i <= n; ++i) {
    for (auto it : g[i])
      if (ctc[it] != ctc[i]) {
        ++d[ctc[it]];
      }
  }
  for (int i = 1; i <= id; ++i)
    if (ctc[s] != i && !d[i]) ++sol;
  cout << sol;
  return 0;
}
