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 n, m, s, a, b;
vector<int> adjlist[5001];
vector<int> jdalist[5001];
vector<pair<int, int> > edgelist;
vector<int> compadj[5001];
stack<int> nodes;
int indegree[5001];
bool vis[5001];
int cnt = 0;
int compval[5001];
void dfs(int node) {
  if (!vis[node]) {
    vis[node] = true;
    for (int i = 0; i < adjlist[node].size(); i++) dfs(adjlist[node][i]);
  }
}
void pushstackdfs(int node) {
  if (!vis[node]) {
    vis[node] = true;
    for (int i = 0; i < adjlist[node].size(); i++)
      pushstackdfs(adjlist[node][i]);
    nodes.push(node);
  }
}
void setdfs(int node, int ccval) {
  if (!vis[node]) {
    vis[node] = true;
    compval[node] = ccval;
    for (int i = 0; i < jdalist[node].size(); i++) {
      setdfs(jdalist[node][i], ccval);
    }
  }
}
void compdfs(int node) {
  if (!vis[node]) {
    vis[node] = true;
    for (int i = 0; i < compadj[node].size(); i++) compdfs(compadj[node][i]);
  }
}
int main() {
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    adjlist[a].push_back(b);
    jdalist[b].push_back(a);
    edgelist.push_back(make_pair(a, b));
  }
  for (int i = 0; i <= n; i++) vis[i] = false;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      pushstackdfs(i);
    }
  }
  for (int i = 0; i <= n; i++) vis[i] = false;
  int cc = 0;
  while (!nodes.empty()) {
    int curnode = nodes.top();
    nodes.pop();
    if (!vis[curnode]) {
      setdfs(curnode, cc);
      cc++;
    }
  }
  for (int i = 0; i < edgelist.size(); i++) {
    if (compval[edgelist[i].first] != compval[edgelist[i].second]) {
      compadj[compval[edgelist[i].first]].push_back(
          compval[edgelist[i].second]);
      indegree[compval[edgelist[i].second]]++;
    }
  }
  for (int i = 0; i < cc; i++) vis[i] = false;
  compdfs(compval[s]);
  for (int i = 0; i < cc; i++) {
    if (indegree[i] == 0 && !vis[i]) {
      compdfs(i);
      cnt++;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      compdfs(i);
      cnt++;
    }
  }
  cout << cnt << '\n';
}
