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 maxn = 5000 + 5;
vector<int> g[maxn], r[maxn], tp;
vector<int> z[maxn];
int sz, scc[maxn];
bool v[maxn];
void dfs(int x) {
  v[x] = true;
  for (int u : g[x])
    if (!v[u]) dfs(u);
  tp.push_back(x);
}
void rdfs(int x) {
  scc[x] = sz;
  for (int u : r[x])
    if (!scc[u]) rdfs(u);
}
bool reach(int x, int t) {
  if (x == t) return true;
  bool f = false;
  v[x] = true;
  for (int u : z[x])
    if (!v[u]) f |= reach(u, t);
  return f;
}
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  --s;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    r[b].push_back(a);
  }
  for (int i = 0; i < n; ++i)
    if (!v[i]) dfs(i);
  reverse(tp.begin(), tp.end());
  for (int i : tp)
    if (!scc[i]) ++sz, rdfs(i);
  for (int i = 0; i < n; ++i) {
    for (int j : g[i])
      if (scc[i] != scc[j]) z[scc[i]].push_back(scc[j]);
  }
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    memset(v, false, sizeof(v));
    if (!reach(scc[s], scc[tp[i]])) z[scc[s]].push_back(scc[tp[i]]), ++ans;
  }
  printf("%d\n", ans);
}
