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 = 2e5 + 10;
const int mod = 1e9 + 7;
vector<int> g[N];
int vis[N], dfn[N], low[N], gcnt, ln, fa[N], p[N];
stack<int> s;
void tarjan(int x) {
  dfn[x] = low[x] = ++gcnt;
  vis[x] = 1;
  s.push(x);
  for (int i = 0; i < g[x].size(); i++) {
    int to = g[x][i];
    if (!dfn[to]) {
      tarjan(to);
      low[x] = min(low[x], low[to]);
    } else if (vis[to])
      low[x] = min(low[x], dfn[to]);
  }
  int cur;
  if (dfn[x] == low[x]) {
    ++ln;
    do {
      cur = s.top();
      s.pop();
      fa[cur] = ln;
      vis[cur] = 0;
    } while (cur != x);
  }
}
int n, m, k;
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  p[fa[k]] = 1;
  int ans = 0;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < g[i].size(); j++) {
      int to = g[i][j];
      if (fa[to] != fa[i]) p[fa[to]]++;
    }
  for (int i = 1; i <= ln; i++)
    if (!p[i]) ans++;
  printf("%d\n", ans);
}
