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 int V = 4e5 + 10;
vector<int> G[V];
vector<int> rG[V];
vector<int> vs, cmp[V];
bool vis[V];
int id[V];
void add_edge(int u, int v) {
  G[u].push_back(v);
  rG[v].push_back(u);
}
void dfs(int u) {
  vis[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    if (!vis[G[u][i]]) dfs(G[u][i]);
  }
  vs.push_back(u);
}
void rdfs(int u, int k) {
  vis[u] = 1;
  cmp[k].push_back(u);
  id[u] = k;
  for (int i = 0; i < rG[u].size(); i++) {
    if (!vis[rG[u][i]]) rdfs(rG[u][i], k);
  }
}
int scc(int n) {
  memset(vis, 0, sizeof(vis));
  vs.clear();
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  memset(vis, 0, sizeof(vis));
  int id = 0;
  for (int i = vs.size() - 1; i >= 0; i--) {
    if (!vis[vs[i]]) rdfs(vs[i], ++id);
  }
  return id;
}
int in[V];
void solve() {
  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);
    add_edge(u, v);
  }
  int n2 = scc(n);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)G[i].size(); j++) {
      int v = G[i][j];
      if (id[i] == id[v]) continue;
      in[id[v]]++;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n2; i++) {
    if (in[i] == 0) ans++;
  }
  if (in[id[s]] == 0) ans--;
  printf("%d\n", ans);
}
int main() {
  solve();
  return 0;
}
