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 = 1e4 + 5;
int head[N], tot;
struct Edge {
  int to, next;
} e[N];
void addEdge(int u, int v) {
  e[tot] = {v, head[u]};
  head[u] = tot++;
}
bool inStack[N];
int dfn[N], low[N], S[N], id[N], top, dfsNum, idx;
void Tarjan(int u) {
  dfn[u] = low[u] = ++dfsNum;
  S[top++] = u;
  inStack[u] = 1;
  for (int i = head[u]; i != -1; i = e[i].next) {
    int v = e[i].to;
    if (!dfn[v]) {
      Tarjan(v);
      low[u] = min(low[u], low[v]);
    } else if (inStack[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (dfn[u] == low[u]) {
    int v;
    do {
      id[v = S[--top]] = idx;
      inStack[v] = 0;
    } while (u != v);
    ++idx;
  }
}
void SCC(int n) {
  dfsNum = top = idx = 0;
  memset(dfn, 0, sizeof(int) * (n + 5));
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) Tarjan(i);
}
int d[N];
int main() {
  int n, m, s;
  scanf("%d%d%d", &n, &m, &s);
  memset(head, -1, sizeof head);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addEdge(u, v);
  }
  SCC(n);
  for (int i = 1; i <= n; i++)
    for (int j = head[i]; j != -1; j = e[j].next)
      if (id[i] != id[e[j].to]) d[id[e[j].to]]++;
  int ans = 0;
  for (int i = 0; i < idx; i++) ans += d[i] == 0 && i != id[s];
  printf("%d\n", ans);
  return 0;
}
