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;
const int MAX = 10000;
int V, E, cnt, dfsn[MAX], SN, sn[MAX];
vector<int> adj[MAX];
bool finished[MAX];
stack<int> S;
vector<vector<int>> SCC;
int DFS(int curr) {
  dfsn[curr] = ++cnt;
  S.push(curr);
  int result = dfsn[curr];
  for (int next : adj[curr]) {
    if (dfsn[next] == 0)
      result = min(result, DFS(next));
    else if (!finished[next])
      result = min(result, dfsn[next]);
  }
  if (result == dfsn[curr]) {
    vector<int> currSCC;
    while (1) {
      int t = S.top();
      S.pop();
      currSCC.push_back(t);
      finished[t] = true;
      sn[t] = SN;
      if (t == curr) break;
    }
    sort(currSCC.begin(), currSCC.end());
    SCC.push_back(currSCC);
    SN++;
  }
  return result;
}
vector<int> sAdj[500000];
int sStart;
int sOutdegree[500000] = {0};
int vist[5001] = {
    0,
};
void dfs2(int num) {
  vist[num] = 1;
  for (auto next : sAdj[num]) {
    if (vist[next] == 0) {
      dfs2(next);
    }
  }
}
int main() {
  int N, M, cash[500000];
  bool hasRest[500000] = {0};
  scanf("%d %d", &N, &M);
  int start;
  scanf("%d", &start);
  for (int i = 0; i < M; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    u--;
    v--;
    adj[u].push_back(v);
  }
  for (int i = 0; i < N; i++)
    if (dfsn[i] == 0) DFS(i);
  start--;
  for (int i = 0; i < N; i++) {
    int si = sn[i];
    if (i == start) sStart = si;
    for (int j : adj[i]) {
      int sj = sn[j];
      if (si == sj) continue;
      sAdj[si].push_back(sj);
      sOutdegree[sj]++;
    }
  }
  dfs2(sStart);
  int anw = 0;
  for (int i = 0; i < SN; i++) {
    if (sOutdegree[i] == 0 && i != sStart) {
      anw++;
    }
  }
  cout << anw;
}
