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 double pi = acos(-1.0);
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
template <typename T>
inline void write(T x) {
  int i = 20;
  char buf[21];
  buf[20] = '\n';
  do {
    buf[--i] = x % 10 + '0';
    x /= 10;
  } while (x);
  do putchar(buf[i]);
  while (buf[i++] != '\n');
}
template <typename T>
inline T readInt() {
  T n = 0, s = 1;
  char p = getchar();
  if (p == '-') s = -1;
  while ((p < '0' || p > '9') && p != EOF && p != '-') p = getchar();
  if (p == '-') s = -1, p = getchar();
  while (p >= '0' && p <= '9') {
    n = (n << 3) + (n << 1) + (p - '0');
    p = getchar();
  }
  return n * s;
}
vector<long long int> *g;
bool *ok, *used;
long long int cnt;
void dfs1(long long int v) {
  ok[v] = true;
  for (auto to : g[v])
    if (!ok[to]) dfs1(to);
}
void dfs2(long long int v) {
  used[v] = true;
  cnt++;
  for (auto to : g[v])
    if (!used[to] && !ok[to]) dfs2(to);
}
int main() {
  long long int n, m, s, ans;
  vector<pair<long long int, long long int>> val;
  n = readInt<long long int>();
  g = new vector<long long int>[n];
  ok = new bool[n];
  used = new bool[n];
  m = readInt<long long int>();
  s = readInt<long long int>();
  s--;
  for (int i = 0; i < m; i += 1) {
    long long int u, v;
    u = readInt<long long int>();
    u--;
    v = readInt<long long int>();
    v--;
    g[u].push_back(v);
  }
  memset(ok, false, n * sizeof(bool));
  dfs1(s);
  for (int i = 0; i < n; i += 1)
    if (!ok[i]) {
      cnt = 0;
      memset(used, false, n * sizeof(bool));
      dfs2(i);
      val.push_back(make_pair(cnt, i));
    }
  sort(val.begin(), val.end());
  reverse(val.begin(), val.end());
  ans = 0;
  for (auto it : val)
    if (!ok[it.second]) {
      ans++;
      dfs1(it.second);
    }
  write(ans);
  return 0;
}
