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 maxn = 5e3 + 10;
int n, m, s;
int head[maxn], hcnt = 0;
struct edge {
  int v, next;
} e[maxn];
void add(int u, int v) {
  e[hcnt] = edge{v, head[u]};
  head[u] = hcnt++;
}
int dfn[maxn], low[maxn], suo[maxn], scnt = 0, tcnt = 0, vis[maxn];
stack<int> sta;
void dfs(int u) {
  dfn[u] = low[u] = ++tcnt;
  sta.push(u);
  vis[u] = 1;
  for (int i = head[u]; i >= 0; i = e[i].next) {
    int v = e[i].v;
    if (!dfn[v]) {
      dfs(v);
      low[u] = min(low[v], low[u]);
    } else if (vis[v])
      low[u] = min(low[u], dfn[v]);
  }
  if (low[u] == dfn[u]) {
    ++scnt;
    int k = sta.top();
    while (k != u) {
      vis[k] = 0;
      suo[k] = scnt;
      sta.pop();
      k = sta.top();
    }
    vis[k] = 0;
    suo[k] = scnt;
    sta.pop();
  }
}
int in[maxn];
int main() {
  memset(head, -1, sizeof(head));
  scanf("%d%d%d", &n, &m, &s);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) dfs(i);
  for (int i = 1; i <= n; i++) {
    for (int j = head[i]; j >= 0; j = e[j].next) {
      int v = e[j].v;
      if (suo[i] != suo[v]) {
        in[suo[v]]++;
      }
    }
  }
  int cnt = 0;
  for (int i = 1; i <= scnt; i++)
    if (in[i] == 0) cnt++;
  if (in[suo[s]] == 0) cnt--;
  printf("%d\n", cnt);
  return 0;
}
