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, s;
vector<int> e[N];
int low[N], dfn[N], idx;
bool vis[N];
stack<int> st;
bool in_stack[N];
int color[N], color_tot;
int indeg[N];
int ans;
void AddEdge(int u, int v) { e[u].push_back(v); }
void Dfs(int u) {
  low[u] = dfn[u] = ++idx;
  in_stack[u] = true;
  st.push(u);
  for (int v : e[u]) {
    if (!dfn[v]) {
      Dfs(v);
      low[u] = min(low[u], low[v]);
    } else if (in_stack[v]) {
      low[u] = min(low[u], dfn[v]);
    }
  }
  if (low[u] == dfn[u]) {
    ++color_tot;
    int v;
    do {
      v = st.top();
      st.pop();
      in_stack[v] = false;
      color[v] = color_tot;
    } while (v != u);
  }
}
void Tarjan() {
  for (int i = 1; i <= n; ++i)
    if (!color[i]) Dfs(i);
}
int main() {
  scanf("%d%d%d", &n, &m, &s);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    AddEdge(u, v);
  }
  Tarjan();
  for (int i = 1; i <= n; ++i)
    ;
  for (int i = 1; i <= n; ++i)
    for (int j : e[i])
      if (color[i] != color[j]) ++indeg[color[j]];
  for (int i = 1; i <= color_tot; ++i)
    ;
  for (int i = 1; i <= color_tot; ++i)
    if (indeg[i] == 0) ++ans;
  if (indeg[color[s]] == 0) --ans;
  printf("%d\n", ans);
  return 0;
}
