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 MOD = 1000000007;
const int M = 1000005;
const int N = 5005;
int n, m, p, i, j, s, tot, ans, low[N], dfn[N], d[N], f[N], fa[N], op, ed, r[N],
    t1, t2;
vector<int> g[N], gg[N];
stack<int> sta;
void tarjan(int x) {
  sta.push(x);
  f[x] = 1;
  dfn[x] = low[x] = ++tot;
  for (auto &i : g[x])
    if (!dfn[i]) {
      tarjan(i);
      low[x] = min(low[x], low[i]);
    } else if (f[i])
      low[x] = min(low[x], dfn[i]);
  if (low[x] == dfn[x]) {
    fa[x] = ++s;
    f[x] = 0;
    while (sta.top() != x) {
      fa[sta.top()] = s;
      f[sta.top()] = 0;
      sta.pop();
    }
    sta.pop();
  }
}
void bfs(int x) {
  memset(f, 0, sizeof(f));
  op = ed = 1;
  d[1] = x;
  f[x] = 1;
  while (op <= ed) {
    for (auto &i : gg[d[op]])
      if (!f[i]) {
        d[++ed] = i;
        f[i] = 1;
      }
    op++;
  }
}
int main() {
  memset(dfn, 0, sizeof(dfn));
  memset(f, 0, sizeof(f));
  memset(r, 0, sizeof(r));
  scanf("%d %d %d", &n, &m, &p);
  for (i = 1; i <= n; i++) fa[i] = i;
  for (i = 1; i <= m; i++) {
    scanf("%d %d", &t1, &t2);
    g[t1].push_back(t2);
  }
  for (i = 1, tot = s = 0; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  for (i = 1; i <= n; i++) {
    for (auto &j : g[i])
      if (fa[j] != fa[i]) {
        gg[fa[i]].push_back(fa[j]);
        r[fa[j]]++;
      }
  }
  bfs(fa[p]);
  for (i = 1, ans = 0; i <= s; i++)
    if (!f[i] && !r[i]) ans++;
  printf("%d\n", ans);
  return 0;
}
