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;
int const MAXn = 5e3 + 2;
vector<int> adya[MAXn];
vector<int> rev[MAXn];
bool mark[MAXn];
int cnt;
int vdfs[MAXn], lowlink[MAXn];
stack<int> pila;
int comp[MAXn];
int cmp;
int ndfs;
void Tarjan(int u) {
  vdfs[u] = lowlink[u] = ++ndfs;
  pila.push(u);
  for (auto to : adya[u]) {
    if (mark[to]) continue;
    if (!vdfs[to]) {
      Tarjan(to);
      lowlink[u] = min(lowlink[u], lowlink[to]);
    } else if (comp[to] == 0)
      lowlink[u] = min(lowlink[u], vdfs[to]);
  }
  if (vdfs[u] == lowlink[u]) {
    cmp++;
    int w;
    do {
      w = pila.top();
      pila.pop();
      comp[w] = cmp;
    } while (vdfs[u] != vdfs[w]);
  }
}
set<int> mapa[MAXn];
bool yep[MAXn];
void DFS(int nod) {
  yep[nod] = true;
  for (auto to : mapa[nod]) {
    if (yep[to]) continue;
    DFS(to);
  }
  if (mapa[nod].empty()) {
    cnt++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int n1, n2;
    cin >> n1 >> n2;
    adya[n2].push_back(n1);
    rev[n1].push_back(n2);
  }
  mark[k] = true;
  queue<int> cola;
  cola.push(k);
  while (!cola.empty()) {
    int v = cola.front();
    cola.pop();
    for (auto to : rev[v]) {
      if (mark[to]) continue;
      cola.push(to);
      mark[to] = true;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (mark[i] || comp[i]) continue;
    Tarjan(i);
  }
  for (int i = 1; i <= n; i++) {
    if (mark[i]) continue;
    for (auto to : adya[i]) {
      if (comp[to] == comp[i]) continue;
      mapa[comp[i]].insert(comp[to]);
    }
  }
  for (int i = 1; i <= cmp; i++) {
    if (yep[i]) continue;
    DFS(i);
  }
  cout << cnt << "\n";
  return 0;
}
