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;
vector<int> vv[5005];
struct node {
  int u, ft;
};
node arr[5005];
int vis[5005];
int T;
void dfs(int u) {
  vis[u] = 1;
  T++;
  for (int i = 0; i < vv[u].size(); i++) {
    int v = vv[u][i];
    if (vis[v] == 0) dfs(v);
  }
  arr[u].ft = ++T;
}
bool comp(node a, node b) { return a.ft > b.ft; }
void dfs2(int u) {
  vis[u] = 1;
  for (int i = 0; i < vv[u].size(); i++) {
    int v = vv[u][i];
    if (vis[v] == 0) dfs2(v);
  }
}
int main() {
  int n, m, s;
  scanf("%d %d %d", &n, &m, &s);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    vv[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) arr[i].u = i;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i);
  }
  sort(arr + 1, arr + n + 1, comp);
  memset(vis, 0, sizeof(vis));
  dfs2(s);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int u = arr[i].u;
    if (vis[u] == 0) {
      ans++;
      dfs2(u);
    }
  }
  cout << ans << endl;
}
