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 start;
void dfs(int par, int node, int vis[], vector<int> v[], int vis1[]) {
  vis[node] = par;
  vis1[node] = 1;
  for (int i = 0; i < v[node].size(); i++) {
    if (vis1[v[node][i]] == 0) dfs(par, v[node][i], vis, v, vis1);
  }
}
int main() {
  int n, m;
  cin >> n >> m >> start;
  vector<int> v[n + 1];
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    v[l].push_back(r);
  }
  int vis[n + 1], count = 0, vis1[n + 1];
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    memset(vis1, 0, sizeof(vis1));
    if (vis[i] == 0) dfs(i, i, vis, v, vis1);
  }
  memset(vis1, 0, sizeof(vis1));
  dfs(start, start, vis, v, vis1);
  map<int, int> map;
  for (int i = 1; i <= n; i++) {
    if (vis1[i] == 0) map[vis[i]]++;
  }
  cout << map.size();
  return 0;
}
