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>
#pragma GCC optimize("Ofast")
using namespace std;
int dx[] = {0, 0, 1, -1, -1, 1, -1, 1};
int dy[] = {-1, 1, 0, 0, 1, 1, -1, -1};
int dz[] = {0, 0, 1, -1, -1, 1, -1, 1};
int kx[] = {2, 2, -2, -2, 1, -1, 1, -1};
int ky[] = {1, -1, 1, -1, 2, 2, -2, -2};
template <typename T>
void print(const T& x) {
  cerr << x << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
bool isPal(string s) {
  int half = s.size() / 2;
  for (int i = 0, j = s.size() - 1; i < half; i++, j--)
    if (s[i] != s[j]) return false;
  return true;
}
double _distance(double x1, double y1, double x2, double y2) {
  double x1x2 = x1 - x2;
  double y1y2 = y1 - y2;
  return sqrt((x1x2 * x1x2) + (y1y2 * y1y2));
}
bool inBound(int x, int y, int r, int c) {
  return (x >= 0 && x < r && y >= 0 && y < c);
}
long long mulmod(long long a, long long b, long long c) {
  long long x = 0, y = a % c;
  while (b > 0) {
    if (b & 1) x = (x + y) % c;
    y = (y << 1) % c;
    b >>= 1;
  }
  return x % c;
}
const int N = 5005;
vector<int> g[N];
stack<int> st;
bool visit[N];
void dfs(int u) {
  visit[u] = true;
  for (auto v : g[u])
    if (!visit[v]) dfs(v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int n, m, s, u, v;
  cin >> n >> m >> s;
  while (m--) {
    cin >> u >> v;
    g[u].push_back(v);
  }
  dfs(s);
  for (int i = 1; i <= n; i++)
    if (!visit[i]) {
      dfs(i);
      st.push(i);
    }
  memset(visit, false, sizeof(visit));
  int road = 0;
  while (st.size()) {
    int u = st.top();
    st.pop();
    if (!visit[u]) {
      dfs(u);
      road++;
    }
  }
  cout << road << endl;
  return 0;
}
