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;
vector<vector<long long>> g, gr;
vector<long long> ord, used, cmp;
void dfs(long long v) {
  used[v] = 1;
  for (auto i : g[v])
    if (!used[i]) dfs(i);
  ord.push_back(v);
}
long long t = 0;
void dfs2(long long v) {
  cmp[v] = t;
  used[v] = 1;
  for (auto i : gr[v])
    if (!used[i]) dfs2(i);
}
int32_t main() {
  long long n, m, s;
  cin >> n >> m >> s;
  s--;
  used.resize(n);
  g.resize(n);
  gr.resize(n);
  cmp.resize(n);
  for (int i = 0; i < m; i++) {
    long long x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    gr[y].push_back(x);
  }
  for (int i = 0; i < n; i++)
    if (!used[i]) dfs(i);
  used.assign(n, 0);
  reverse(ord.begin(), ord.end());
  for (auto i : ord)
    if (!used[i]) {
      dfs2(i);
      t++;
    }
  set<long long> stok;
  for (int i = 0; i < n; i++) {
    bool l = 0;
    for (auto j : g[i])
      if (cmp[i] != cmp[j]) {
        stok.insert(cmp[j]);
      }
  }
  set<long long> ans;
  for (int i = 0; i < n; i++)
    if (stok.find(cmp[i]) != stok.end()) ans.insert(cmp[i]);
  cout << t - ans.size() - (stok.find(cmp[s]) == stok.end());
  return 0;
}
