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;
const long long INF = 1e18;
const long long SIZE = 5e3 + 5;
const long long mod = 1e9 + 7;
bool ok[SIZE];
bool vis[SIZE];
vector<long long> adj[SIZE];
vector<pair<long long, long long> > conn;
long long cap, cnt;
void dfs(long long u) {
  if (ok[u]) return;
  ok[u] = 1;
  long long i, v;
  for (i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    dfs(v);
  }
}
void dfs1(long long u) {
  if (vis[u]) return;
  vis[u] = 1;
  cnt++;
  long long i, v;
  for (i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    dfs1(v);
  }
}
bool cmp(pair<long long, long long> a, pair<long long, long long> b) {
  return (a.second > b.second);
}
int main() {
  long long t = 0, n, x, y, z, i, j, k, g, p, q, ans = 0, sum = 0, c = 0;
  scanf("%lld %lld", &n, &q);
  scanf("%lld", &cap);
  for (i = 1; i <= q; i++) {
    scanf("%lld %lld", &x, &y);
    adj[x].push_back(y);
  }
  dfs(cap);
  for (i = 1; i <= n; i++) {
    if (!ok[i]) {
      cnt = 0;
      memset(vis, 0, sizeof vis);
      dfs1(i);
      conn.push_back(make_pair(i, cnt));
    }
  }
  sort(conn.begin(), conn.end(), cmp);
  for (i = 0; i < conn.size(); i++) {
    x = conn[i].first;
    if (!ok[x]) {
      dfs(x);
      ans++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
