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 INF = 1e9;
const int MAX = 1e5 + 7, N = 1e5 + 3;
double pi = cos(-1);
int r, l, k, m, n, ans;
bool visited1[10000];
bool visited2[10000];
vector<pair<int, int> > uv;
int h[10000];
vector<vector<int> > graph;
int s;
list<int> bfslist;
int sour;
bool done;
void bfs(int v) {
  for (int i = 0; i < graph[v].size(); i++) {
    if (visited1[graph[v][i]] or visited2[graph[v][i]]) continue;
    bfslist.push_back(graph[v][i]);
    visited1[graph[v][i]] = true;
    done = true;
  }
}
void bfs2(int v) {
  for (int i = 0; i < graph[v].size(); i++) {
    if (visited1[graph[v][i]] or visited2[graph[v][i]]) continue;
    bfslist.push_back(graph[v][i]);
    visited2[graph[v][i]] = true;
    h[sour]++;
  }
}
int main() {
  cin >> n >> m >> s;
  graph.assign(n + 2, vector<int>());
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
  }
  bfslist.push_back(s);
  visited1[s] = true;
  while (!bfslist.empty()) {
    int v = bfslist.front();
    bfslist.pop_front();
    bfs(v);
  }
  for (int i = 1; i <= n; i++) {
    if (visited1[i]) continue;
    bfslist.push_back(i);
    memset(visited2, 0, sizeof(visited2));
    visited2[i] = true;
    sour = i;
    while (!bfslist.empty()) {
      int v = bfslist.front();
      bfslist.pop_front();
      bfs2(v);
    }
    uv.push_back(make_pair(-1 * h[sour], sour));
  }
  memset(visited2, 0, sizeof(visited2));
  sort(uv.begin(), uv.end());
  for (int i = 0; i < uv.size(); i++) {
    if (visited1[uv[i].second]) continue;
    bfslist.push_back(uv[i].second);
    visited1[uv[i].second] = true;
    while (!bfslist.empty()) {
      int v = bfslist.front();
      bfslist.pop_front();
      bfs(v);
    }
    ans++;
  }
  cout << ans << endl;
  return 0;
}
