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 = 5005;
vector<int> v[N], vt[N], scc[N];
bool viz[N];
stack<int> comp;
int nr, from[N], gr[N];
void dfp(int i) {
  viz[i] = 1;
  for (auto it : v[i])
    if (!viz[it]) dfp(it);
  comp.push(i);
}
void dfm(int i) {
  scc[nr].push_back(i);
  viz[i] = 1;
  for (auto it : vt[i])
    if (!viz[it]) dfm(it);
}
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    vt[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (!viz[i]) dfp(i);
  memset(viz, 0, sizeof(viz));
  while (!comp.empty()) {
    int now = comp.top();
    comp.pop();
    if (!viz[now]) {
      nr++;
      dfm(now);
    }
  }
  int ans = 0;
  for (int i = 1; i <= nr; i++)
    for (auto it : scc[i]) from[it] = i;
  for (int i = 1; i <= n; i++)
    for (auto it : v[i])
      if (from[it] != from[i]) gr[from[it]]++;
  for (int i = 1; i <= nr; i++) {
    bool ok = 1;
    for (auto it : scc[i])
      if (it == s) ok = 0;
    if (ok && gr[i] == 0) ans++;
  }
  cout << ans;
}
