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_N = 5000 + 10;
int n, m, s;
vector<int> G[MAX_N];
bool reach[MAX_N], used[MAX_N];
void dfs(int v) {
  used[v] = true;
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    if (!used[u]) dfs(u);
  }
}
void dfs2(int v) {
  reach[v] = true;
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    if (!reach[u]) dfs2(u);
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &s);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].push_back(v);
  }
  memset(used, false, sizeof(used));
  dfs(s);
  for (int v = 1; v <= n; v++) {
    if (!used[v]) {
      memset(reach, 0, sizeof(reach));
      dfs2(v);
      reach[v] = false;
      for (int i = 1; i <= n; i++) {
        if (reach[i]) used[i] = true;
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!used[i]) {
      ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}
