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;
int done[5001] = {0};
vector<int> adj[5001];
int income[5001][5001];
long long ans = 0;
stack<int> stack1;
void DFSUtil1(long long u, vector<bool> &visited, long long p) {
  visited[u] = true;
  long long i;
  for (i = 0; i < adj[u].size(); i++)
    if (visited[adj[u][i]] == false && done[adj[u][i]] == 0)
      DFSUtil1(adj[u][i], visited, u);
  stack1.push(u);
}
void DFSUtil2(long long u, vector<long long> adj[], vector<bool> &visited,
              long long p, vector<long long> *temp) {
  visited[u] = true;
  temp->push_back(u);
  long long i;
  for (i = 0; i < adj[u].size(); i++)
    if (visited[adj[u][i]] == false && done[adj[u][i]] == 0)
      DFSUtil2(adj[u][i], adj, visited, u, temp);
}
void SCC(long long s) {
  long long u, i, j;
  vector<bool> visited(s, false);
  for (u = 0; u < s; u++)
    if (visited[u] == false && done[u] == 0) DFSUtil1(u, visited, -1);
  vector<long long> trps[s];
  for (i = 0; i < s; i++) {
    for (j = 0; j < adj[i].size(); j++) {
      trps[adj[i][j]].push_back(i);
    }
  }
  for (i = 0; i < visited.size(); i++) visited[i] = false;
  while (!stack1.empty()) {
    long long v = stack1.top();
    stack1.pop();
    if (visited[v] == false && done[v] == 0) {
      vector<long long> *temp = new vector<long long>;
      vector<long long> in;
      DFSUtil2(v, trps, visited, -1, temp);
      for (i = 0; i < temp->size(); i++) {
        for (j = 0; j < temp->size(); j++) income[(*temp)[i]][(*temp)[j]] = 0;
      }
      int l = 0;
      for (i = 0; i < temp->size(); i++) {
        for (j = 0; j < 5001; j++) {
          if (income[(*temp)[i]][j] == 1) l = 1;
        }
      }
      if (l == 0) ans++;
    }
  }
}
void dfs(int n) {
  done[n] = 1;
  for (int i = 0; i < adj[n].size(); i++) {
    if (done[adj[n][i]] != 1) dfs(adj[n][i]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, s, b, c, i, j, k, l, m, t, x, y, n, z;
  cin >> n >> m >> s;
  for (i = 0; i < 5001; i++)
    for (j = 0; j < 5001; j++) income[i][j] = 0;
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    income[y][x] = 1;
  }
  s--;
  dfs(s);
  SCC(n);
  cout << ans << endl;
}
