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;
long long int MOD(long long int a, long long int b) {
  if (a > b)
    return a - b;
  else
    return b - a;
}
int max3(int a, int b, int c) { return max(c, max(a, b)); }
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int logg(long long int a) {
  long long int x = 0;
  while (a > 1) {
    x++;
    a /= 2;
  }
  return x;
}
vector<int> adj[200005], visited;
vector<int> B;
void dfs(int x) {
  visited[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    if (!visited[adj[x][i]]) dfs(adj[x][i]);
  }
  B.push_back(x);
}
void dfs1(int x) {
  visited[x] = 1;
  for (int i = 0; i < adj[x].size(); i++) {
    if (!visited[adj[x][i]]) dfs1(adj[x][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, m, S;
  cin >> n >> m >> S;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
  }
  visited.resize(n + 2, 0);
  dfs1(S);
  for (int i = 1; i <= n; i++)
    if (!visited[i]) dfs(i);
  visited.clear();
  visited.resize(n + 2, 0);
  int x = 0;
  dfs1(S);
  for (int i = B.size() - 1; i >= 0; i--) {
    if (!visited[B[i]]) {
      dfs1(B[i]);
      x++;
    }
  }
  cout << x;
}
