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;
using LL = int64_t;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const LL LLINF = 1e14;
const double pi = acos(-1.0);
vector<int> E[maxn];
bool vis[maxn], tvis[maxn];
void dfs1(int x) {
  vis[x] = true;
  for (int i = 0; i < E[x].size(); i++) {
    int v = E[x][i];
    if (vis[v] == true) continue;
    dfs1(v);
  }
  return;
}
void dfs2(int x) {
  vis[x] = 1;
  for (int i = 0; i < E[x].size(); i++) {
    int v = E[x][i];
    if (vis[v]) continue;
    if (tvis[v]) tvis[v] = false;
    dfs2(v);
  }
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m, s;
  cin >> n >> m >> s;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    E[x].push_back(y);
  }
  dfs1(s);
  for (int i = 1; i <= n; i++)
    if (vis[i] == false) tvis[i] = true;
  for (int i = 1; i <= n; i++) {
    if (tvis[i] == true) {
      memset(vis, false, sizeof(vis));
      dfs2(i);
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (tvis[i] == true) ans++;
  }
  cout << ans;
  return 0;
}
