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;
vector<int> g[(int)1e4 + 10];
int dfn[(int)1e4 + 10], low[(int)1e4 + 10], col[(int)1e4 + 10],
    s[(int)1e4 + 10];
bool visit[(int)1e4 + 10];
int cnt, top, idx;
void tarjan(int now) {
  dfn[now] = low[now] = ++cnt;
  visit[now] = true;
  s[++top] = now;
  int i;
  for (i = 0; i < g[now].size(); i++) {
    int temp = g[now][i];
    if (!dfn[temp]) {
      tarjan(temp);
      low[now] = min(low[now], low[temp]);
    } else if (visit[temp]) {
      low[now] = min(low[now], dfn[temp]);
    }
  }
  if (low[now] == dfn[now]) {
    idx++;
    while (s[top] != now) {
      visit[s[top]] = false;
      col[s[top]] = idx;
      top--;
    }
    visit[now] = false;
    col[now] = idx;
    top--;
  }
}
int main(void) {
  int n, m, st;
  scanf("%d%d%d", &n, &m, &st);
  int i, j;
  for (i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
  }
  for (i = 1; i <= n; i++) {
    if (!dfn[i]) tarjan(i);
  }
  int d[(int)1e4 + 10];
  for (i = 1; i <= n; i++) {
    for (j = 0; j < g[i].size(); j++) {
      if (col[i] == col[g[i][j]]) continue;
      d[col[g[i][j]]]++;
    }
  }
  int res = 0;
  for (i = 1; i <= idx; i++) {
    if (d[i] == 0 && i != col[st]) res++;
  }
  printf("%d", res);
}
