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;
struct node {
  int id, cnt;
} A[5000 + 5];
int tmp;
bool vis[2][5000 + 5];
vector<int> G[5000 + 5];
bool cmp(node x, node y) { return x.cnt > y.cnt; }
void dfs(int u, int id) {
  tmp++;
  vis[id][u] = 1;
  for (int i = 0; i < int(G[u].size()); i++) {
    int v = G[u][i];
    if (!vis[id][v]) dfs(v, id);
  }
}
int main() {
  scanf("%d%d%d", &N, &M, &S);
  for (int i = 1; i <= M; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
  }
  dfs(S, 0);
  int Ans = 0;
  for (int i = 1; i <= N; i++)
    if (!vis[0][i]) {
      tmp = 0;
      memcpy(vis[1], vis[0], sizeof vis[0]);
      dfs(i, 1);
      A[i].id = i;
      A[i].cnt = tmp;
    }
  sort(A + 1, A + N + 1, cmp);
  for (int i = 1; A[i].cnt; i++)
    if (!vis[0][A[i].id]) {
      Ans++;
      dfs(A[i].id, 0);
    }
  printf("%d", Ans);
}
