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, c;
int h[N], ne[N], e[N], idx;
int low[N], dfn[N], timestamp;
int din[N], scc_cnt, id[N];
int stk[N], top;
bool in_stk[N];
void add(int a, int b) { ne[idx] = h[a], e[idx] = b, h[a] = idx++; }
void tarjan(int u) {
  low[u] = dfn[u] = ++timestamp;
  stk[++top] = u, in_stk[u] = true;
  for (int i = h[u]; i != -1; i = ne[i]) {
    int j = e[i];
    if (!dfn[j]) {
      tarjan(j);
      low[u] = min(low[u], low[j]);
    } else if (in_stk[j])
      low[u] = min(low[u], dfn[j]);
  }
  if (dfn[u] == low[u]) {
    ++scc_cnt;
    int y;
    do {
      y = stk[top--];
      in_stk[y] = false;
      id[y] = scc_cnt;
    } while (y != u);
  }
}
int main() {
  memset(h, -1, sizeof(h));
  cin >> n >> m >> c;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    add(a, b);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  for (int i = 1; i <= n; i++)
    for (int j = h[i]; j != -1; j = ne[j]) {
      int k = e[j];
      int a = id[i], b = id[k];
      if (a != b) din[b]++;
    }
  int ans = 0;
  for (int i = 1; i <= scc_cnt; i++)
    if (!din[i]) ans++;
  if (din[id[c]] == 0) ans--;
  cout << ans << endl;
  return 0;
}
