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 ll = long long;
using namespace std;
const ll mod = 1e9 + 7;
const int N = 4005;
int n, m, s, c, c1;
vector<vector<int>> gp;
vector<bool> vis;
vector<int> tmp;
void dfs(int u) {
  vis[u] = 1;
  tmp.push_back(u);
  c++;
  c1++;
  for (int& v : gp[u])
    if (!vis[v]) dfs(v);
}
int main() {
  scanf("%d%d%d", &n, &m, &s);
  s--;
  gp = vector<vector<int>>(n);
  vis = vector<bool>(n);
  while (m--) {
    int u, v;
    scanf("%d%d", &u, &v);
    gp[--u].push_back(--v);
  }
  c = 0, c1 = 0;
  dfs(s);
  int ans = 0;
  while (c != n) {
    ans++;
    int mx = -1, u = -1;
    for (int i = 0; i < n; i++) {
      if (!vis[i]) {
        c1 = 0;
        tmp.clear();
        dfs(i);
        if (c1 > mx) mx = c1, u = i;
        for (int& j : tmp) vis[j] = 0, c--;
      }
    }
    dfs(u);
  }
  printf("%d", ans);
  return 0;
}
