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 V, E, g[5003][5003];
int cnt, Stack[5003], top;
char vis[5003], ins[5003], ans[5003];
int dfn[5003], low[5003];
int uf[5003];
int used[5003];
struct Edge {
  int u, v;
  Edge(void) {}
  Edge(int uu, int vv) : u(uu), v(vv) {}
} Es[5003];
int find(int x) {
  if (uf[x] >= 0) return uf[x] = find(uf[x]);
  return x;
}
void merge(int x, int y) {
  int r1 = find(x);
  int r2 = find(y);
  if (r1 < r2) {
    uf[r1] += uf[r2];
    uf[r2] = r1;
  } else if (r2 < r1) {
    uf[r2] += uf[r1];
    uf[r1] = r2;
  }
}
void Tarjian(int u) {
  vis[u] = ins[u] = 1;
  dfn[u] = low[u] = ++cnt;
  Stack[top++] = u;
  for (int v = 1; v <= V; ++v) {
    if (g[u][v]) {
      if (!vis[v]) {
        Tarjian(v);
        if (low[u] > low[v]) low[u] = low[v];
      } else if (ins[v]) {
        if (low[u] > dfn[v]) low[u] = dfn[v];
      }
    }
  }
  if (dfn[u] == low[u]) {
    int now, sum = 0;
    while (1) {
      now = Stack[--top];
      ins[now] = 0;
      if (now == u) break;
      merge(now, u);
    }
  }
}
int main() {
  int u, v, e, ccnt = 0, tp, src;
  memset(uf, -1, sizeof(uf));
  scanf("%d %d %d", &V, &E, &src);
  for (e = 0; e < E; ++e) {
    scanf("%d %d", &u, &v);
    g[u][v] = 1;
    Es[e] = Edge(u, v);
  }
  for (u = 1; u <= V; ++u)
    if (!vis[u]) Tarjian(u);
  for (u = 1; u <= V; ++u) tp = find(u), used[tp] = tp;
  for (e = 0; e < E; ++e) {
    tp = find(Es[e].v);
    if (find(Es[e].u) != tp) ans[tp] = 1;
  }
  ans[0] = 1;
  for (u = 1; u <= V; ++u)
    if (!ans[used[u]]) ccnt++;
  if (!ans[find(src)]) ccnt--;
  printf("%d\n", ccnt);
}
