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 = 1e5 + 10;
inline int read() {
  int d = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') d = (d << 1) + (d << 3) + c - '0', c = getchar();
  return d * f;
}
vector<int> adj[N], v, st, tu[N];
int id[N], low[N], pp, kk, in[N], mp[N], ru[N], chu[N], vst[N], ttt;
void dfs(int u) {
  ++ttt;
  st.push_back(u);
  in[u] = 1;
  low[u] = id[u] = ++pp;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (!id[v]) dfs(v);
    if (in[v]) low[u] = min(low[u], low[v]);
  }
  if (low[u] == id[u]) {
    mp[u] = ++kk;
    while (st.back() != u) {
      mp[st.back()] = kk;
      in[st.back()] = 0;
      st.pop_back();
    }
    in[u] = 0;
    st.pop_back();
  }
}
int main() {
  int n = read(), m = read(), s = read();
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read();
    adj[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    if (!id[i]) dfs(i);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < adj[i].size(); j++) {
      int v = adj[i][j];
      if (mp[i] != mp[v]) {
        tu[mp[i]].push_back(mp[v]);
        ru[mp[v]]++, chu[mp[i]]++;
      }
    }
  }
  int ct1 = 0, ct2 = 0;
  queue<int> q;
  for (int i = 1; i <= kk; i++) {
    if (!ru[i] && i != mp[s]) {
      ct1++;
    }
  }
  cout << ct1 << endl;
  return 0;
}
