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 n, m, s;
int num[30];
char a[1001000], b[1001000];
class Tarjan {
 public:
  int s_num;
  int s_in[1001000], s_out[1001000];
  int s_number[1001000];
  vector<int> s_node[1001000];
  void init(int node_num = 0) {
    num = node_num;
    for (int i = 1; i <= num; i++) node[i].clear();
  }
  void addEdge(int u, int v, bool isTurn = true) {
    node[u].push_back(v);
    if (isTurn) node[v].push_back(u);
  }
  void tarjan() {
    s_num = step = 0;
    memset(dfn, 0, sizeof dfn);
    memset(low, 0, sizeof low);
    memset(s_in, 0, sizeof s_in);
    memset(s_out, 0, sizeof s_out);
    memset(node_in_stack, 0, sizeof node_in_stack);
    while (!ns.empty()) ns.pop();
    for (int i = 1; i <= num; i++)
      if (!dfn[i]) tarjan_dfs(i);
    for (int i = 1; i <= s_num; i++) s_node[i].clear();
    for (int i = 1; i <= n; i++) {
      for (int j = 0, len = node[i].size(); j < len; j++) {
        if (s_number[i] != s_number[node[i][j]]) {
          addEdge_S(s_number[i], s_number[node[i][j]]);
          ++s_out[s_number[i]];
          ++s_in[s_number[node[i][j]]];
        }
      }
    }
  }
  void debug() {}

 private:
  int num, step;
  int dfn[1001000], low[1001000];
  bool node_in_stack[1001000];
  stack<int> ns;
  vector<int> node[1001000];
  void addEdge_S(int u, int v) { s_node[u].push_back(v); }
  void tarjan_dfs(int x) {
    dfn[x] = low[x] = ++step;
    ns.push(x);
    node_in_stack[x] = true;
    for (int i = 0, len = node[x].size(); i < len; ++i) {
      int y = node[x][i];
      if (!dfn[y])
        tarjan_dfs(y), low[x] = min(low[x], low[y]);
      else if (node_in_stack[y])
        low[x] = min(low[x], dfn[y]);
    }
    if (dfn[x] == low[x]) {
      ++s_num;
      while (node_in_stack[x]) {
        int y = ns.top();
        ns.pop();
        node_in_stack[y] = false;
        s_number[y] = s_num;
      }
    }
  }
} tarjan;
int main() {
  while (~scanf("%d%d%d", &n, &m, &s)) {
    int ans = 0;
    tarjan.init(n);
    for (int i = 0, x, y; i < m; i++) {
      scanf("%d%d", &x, &y);
      tarjan.addEdge(x, y, false);
    }
    tarjan.tarjan();
    for (int i = 1; i <= tarjan.s_num; i++) {
      if (!tarjan.s_in[i] && i != tarjan.s_number[s]) ans++;
    }
    printf("%d\n", ans);
  }
  return 0;
}
