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 maxn = 2e5 + 3;
const int maxm = 1e7 + 3;
const int mod = 1e9 + 7;
const int oo = 1e9 + 5;
const long double pi = 3.141592653589793238462643383279502884;
void input() {
  if (fopen("abc"
            ".inp",
            "r")) {
    freopen(
        "abc"
        ".inp",
        "r", stdin);
    freopen(
        "abc"
        ".out",
        "w", stdout);
  }
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int n, m, nroot = 0, capital, _time = 0;
vector<int> gb[maxn], gf[maxn];
stack<int> s;
int low[maxn], num[maxn], adj[maxn], vis[maxn], root[maxn];
vector<int> no_path;
void dfs(int u) {
  vis[u] = 1;
  for (int v : gf[u])
    if (!vis[v]) dfs(v);
}
void visit(int u) {
  s.push(u);
  low[u] = num[u] = ++_time;
  for (int v : gb[u]) {
    if (num[v])
      low[u] = min(low[u], num[v]);
    else {
      visit(v);
      low[u] = min(low[u], low[v]);
    }
  }
  if (num[u] == low[u]) {
    int v;
    nroot++;
    do {
      v = s.top();
      s.pop();
      low[v] = num[v] = oo;
      root[v] = nroot;
    } while (v != u);
  }
}
int main() {
  input();
  cin >> n >> m >> capital;
  for (int i = 1, u, v; i <= m; ++i) {
    cin >> u >> v;
    gf[u].push_back(v);
    gb[v].push_back(u);
  }
  dfs(capital);
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) no_path.push_back(i);
  for (int i : no_path)
    if (!num[i]) visit(i);
  for (int u : no_path) {
    for (int v : gb[u])
      if (root[v] != root[u]) adj[root[u]]++;
  }
  int ans = 0;
  for (int i = 1; i <= nroot; ++i) ans += (adj[i] == 0);
  cout << ans;
  return 0;
}
