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 MAX_N = 5015;
vector<int> graph[MAX_N];
vector<int> visited(MAX_N, false);
vector<int> visit_order;
vector<int> reverse_graph[MAX_N];
vector<int> component(MAX_N, 0);
vector<int> component_graph[MAX_N];
vector<int> reachable_component(MAX_N, false);
void dfs(int v) {
  visited[v] = true;
  for (int i = 0; i < graph[v].size(); i++) {
    int child = graph[v][i];
    if (!visited[child]) dfs(child);
  }
  visit_order.push_back(v);
}
void reverse_dfs(int v, int c) {
  component[v] = c;
  for (int i = 0; i < reverse_graph[v].size(); i++) {
    int child = reverse_graph[v][i];
    if (component[child] == 0) reverse_dfs(child, c);
  }
}
void component_dfs(int v) {
  reachable_component[v] = true;
  for (int i = 0; i < component_graph[v].size(); i++) {
    int child_v = component_graph[v][i];
    if (!reachable_component[child_v]) dfs(child_v);
  }
}
int main() {
  int no_of_vertices, no_of_edges, source;
  scanf("%d %d %d", &no_of_vertices, &no_of_edges, &source);
  for (int i = 1; i <= no_of_edges; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    graph[u].push_back(v);
    reverse_graph[v].push_back(u);
  }
  for (int i = 1; i <= no_of_vertices; i++) {
    if (!visited[i]) {
      dfs(i);
    }
  }
  int no_of_components = 0;
  reverse((visit_order).begin(), (visit_order).end());
  for (int i = 0; i < no_of_vertices; i++) {
    int v = visit_order[i];
    if (component[v] == 0) {
      reverse_dfs(v, ++no_of_components);
    }
  }
  vector<int> incoming(no_of_components + 1);
  for (int v = 1; v <= no_of_vertices; v++) {
    for (int i = 0; i < graph[v].size(); i++) {
      int child = graph[v][i];
      if (component[v] != component[child]) {
        component_graph[component[v]].push_back(component[child]);
        incoming[component[child]]++;
      }
    }
  }
  component_dfs(component[source]);
  int new_edges = 0;
  for (int i = 1; i <= no_of_components; i++) {
    if (!reachable_component[i] && incoming[i] == 0) {
      new_edges++;
    }
  }
  printf("%d\n", new_edges);
  return 0;
}
