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 N = 5020;
int nex[N], head[N], to[N];
int dfn[N], low[N], tim, tp, scc, cnt;
int sta[N], ins[N], rd[N], inscc[N];
void add(int x, int y) {
  nex[++cnt] = head[x];
  head[x] = cnt;
  to[cnt] = y;
}
void tarjan(int u) {
  dfn[u] = low[u] = ++tim;
  sta[++tp] = u;
  ins[u] = 1;
  for (int i = head[u]; i; i = nex[i]) {
    int v = to[i];
    if (!dfn[v]) {
      tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (ins[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (low[u] == dfn[u]) {
    ++scc;
    int k;
    do {
      k = sta[tp--];
      inscc[k] = scc;
      ins[k] = 0;
    } while (k != u);
  }
}
int main() {
  int n, m, s, x, y;
  scanf("%d%d%d", &n, &m, &s);
  for (int i = 1; i <= m; i++) scanf("%d%d", &x, &y), add(x, y);
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  for (int i = 1; i <= n; i++)
    for (int j = head[i]; j; j = nex[j])
      if (inscc[i] != inscc[to[j]]) rd[inscc[to[j]]]++;
  int ans = 0;
  for (int i = 1; i <= scc; i++)
    if (!rd[i] && i != inscc[s]) ans++;
  printf("%d", ans);
  return 0;
}
