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 dx4[] = {-1, 0, 1, 0};
int dy4[] = {0, 1, 0, -1};
int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const double eps = (double)1e-10;
const double PI = acos(-1.0);
inline int getint() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long getlonglong() {
  long long x;
  scanf("%lld", &x);
  return x;
}
int n, m, s;
vector<int> graph[5007];
bool vis[5007], visit[5007];
int cnt = 0;
void dfs(int u) {
  if (vis[u]) return;
  cnt++;
  vis[u] = 1;
  int len = graph[u].size();
  for (int i = 0; i < len; i++) {
    int v = graph[u][i];
    if (!vis[v]) dfs(v);
  }
}
void dfs_s(int u) {
  if (visit[u]) return;
  visit[u] = 1;
  int len = graph[u].size();
  for (int i = 0; i < len; i++) {
    int v = graph[u][i];
    if (!visit[v]) dfs_s(v);
  }
}
int main() {
  n = getint(), m = getint(), s = getint();
  for (int i = 0; i < m; i++) {
    int u = getint(), v = getint();
    graph[u].push_back(v);
  }
  dfs_s(s);
  priority_queue<pair<int, int> > pq;
  for (int i = 1; i <= n; i++) {
    if (i == s) continue;
    cnt = 0;
    memset(vis, 0, sizeof vis);
    if (!visit[i]) dfs(i);
    pq.push(pair<int, int>(cnt, i));
  }
  cnt = 0;
  while (!pq.empty()) {
    int u = pq.top().second;
    if (!visit[u]) {
      dfs_s(u);
      cnt++;
    }
    pq.pop();
  }
  printf("%d\n", cnt);
  return 0;
}
