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 = 5010;
int n, m, s;
vector<int> g[N];
vector<int> tg[N];
vector<int> cg[N];
vector<int> ord;
int indeg[N];
bool ucomp[N];
bool used[N];
int comp[N];
int cnt;
void dfs1(int v) {
  used[v] = true;
  for (auto to : g[v])
    if (!used[to]) dfs1(to);
  ord.push_back(v);
}
void dfs2(int v) {
  comp[v] = cnt;
  for (auto to : tg[v])
    if (comp[to] == -1) dfs2(to);
}
void dfs3(int v) {
  ucomp[v] = true;
  for (auto to : cg[v])
    if (!ucomp[to]) dfs3(to);
}
int main() {
  cin >> n >> m >> s;
  --s;
  for (int i = 0; i < m; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    tg[y].push_back(x);
  }
  for (int i = 0; i < n; ++i) {
    if (!used[i]) {
      dfs1(i);
    }
  }
  reverse(ord.begin(), ord.end());
  memset(comp, -1, sizeof(comp));
  for (int i = 0; i < n; ++i) {
    if (comp[ord[i]] == -1) {
      dfs2(ord[i]);
      ++cnt;
    }
  }
  for (int v = 0; v < n; ++v) {
    for (auto to : g[v]) {
      if (comp[v] != comp[to]) {
        cg[comp[v]].push_back(comp[to]);
        ++indeg[comp[to]];
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < cnt; i++) {
    if (!indeg[i] && i != comp[s]) {
      ans++;
    }
  }
  cout << ans << endl;
  return 0;
}
