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;
void err() {
  cout << "shit" << endl;
  exit(1);
}
const int maxn = 5000 + 10;
vector<int> G[2][maxn];
vector<int> out;
pair<int, int> edges[maxn];
int indeg[maxn];
int fa[maxn];
int findset(int x) { return fa[x] == x ? x : fa[x] = findset(fa[x]); }
bool vis[maxn];
void dfs0(int u) {
  vis[u] = true;
  int size = G[0][u].size();
  for (int i = 0; i < size; i++) {
    int v = G[0][u][i];
    if (vis[v]) continue;
    dfs0(v);
  }
  out.push_back(u);
}
void dfs1(int u) {
  vis[u] = true;
  int size = G[1][u].size();
  for (int i = 0; i < size; i++) {
    int v = G[1][u][i];
    if (vis[v]) continue;
    fa[v] = u;
    dfs1(v);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    G[0][u].push_back(v);
    G[1][v].push_back(u);
    edges[i] = make_pair(u, v);
  }
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs0(i);
  for (int i = 1; i <= n; i++) fa[i] = i;
  memset(vis, false, sizeof(vis));
  for (int i = n - 1; i >= 0; i--) {
    int u = out[i];
    if (vis[u]) continue;
    dfs1(u);
  }
  memset(indeg, 0, sizeof(indeg));
  for (int i = 0; i < m; i++) {
    int u = edges[i].first, v = edges[i].second;
    int fu = findset(u), fv = findset(v);
    if (fu != fv) {
      indeg[fv]++;
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++)
    if (findset(i) == i && findset(s) != i && indeg[i] == 0) ans++;
  cout << ans << '\n';
}
