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 u[5005] = {};
bool v[5005] = {};
int n, m, s;
vector<int> V[5005];
void dfs(int now) {
  u[now] = v[now] = 1;
  for (auto x : V[now]) {
    if (v[x]) continue;
    dfs(x);
  }
}
int main() {
  cin >> n >> m >> s;
  s--;
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d %d", &a, &b);
    ;
    a--;
    b--;
    V[a].push_back(b);
  }
  dfs(s);
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (!u[i]) {
      memset(v, 0, sizeof(v));
      dfs(i);
      u[i] = 2;
    }
  }
  for (int i = 0; i < n; ++i)
    if (u[i] == 2) cnt++;
  cout << cnt;
}
