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;
vector<vector<int> > v;
vector<bool> visited;
vector<bool> visited2;
vector<pair<int, vector<int> > > topoSort;
void tSort(int node, int x) {
  if (visited[node] || visited2[node]) return;
  visited[node] = 1;
  if (x == s) visited2[node] = 1;
  for (auto i : v[node]) tSort(i, x);
  topoSort[x].second.push_back(node);
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m >> s;
  v = vector<vector<int> >(n + 1);
  visited = vector<bool>(n + 1);
  visited2 = vector<bool>(n + 1);
  topoSort = vector<pair<int, vector<int> > >(n + 1);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
  }
  tSort(s, s);
  for (int i = 1; i <= n; i++) {
    visited = vector<bool>(n + 1, 0);
    tSort(i, i);
  }
  for (int i = 1; i <= n; i++)
    reverse(topoSort[i].second.begin(), topoSort[i].second.end());
  for (int i = 1; i <= n; i++) topoSort[i].first = topoSort[i].second.size();
  sort(topoSort.rbegin(), topoSort.rend());
  int counter = 0;
  for (int i = 0; i < n; i++) {
    bool c = 0;
    for (auto j : topoSort[i].second) {
      if (!visited2[j]) c = 1;
      visited2[j] = 1;
    }
    if (c) counter++;
  }
  cout << counter;
  return 0;
}
