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;
struct edge {
  int v, next;
} E[5010], nE[5010];
int p[5010], eid;
int np[5010], neid;
void init() {
  memset(p, -1, sizeof(p));
  eid = 0;
}
void ninit() {
  memset(np, -1, sizeof(np));
  neid = 0;
}
void insert(int u, int v) {
  E[eid].v = v;
  E[eid].next = p[u];
  p[u] = eid++;
}
void ninsert(int u, int v) {
  nE[neid].v = v;
  nE[neid].next = np[u];
  np[u] = neid++;
}
int n, m, s;
int dfn[5010], low[5010], belong[5010], cnt = 0, times = 0;
stack<int> sta;
void dfs(int u) {
  dfn[u] = low[u] = ++times;
  sta.push(u);
  for (int i = p[u]; i != -1; i = E[i].next) {
    int v = E[i].v;
    if (dfn[v] == 0) {
      dfs(v);
      low[u] = min(low[u], low[v]);
    } else if (!belong[v]) {
      low[u] = min(low[u], dfn[v]);
    }
  }
  if (low[u] == dfn[u]) {
    ++cnt;
    while (true) {
      int x = sta.top();
      sta.pop();
      belong[x] = cnt;
      if (x == u) {
        break;
      }
    }
  }
}
bool vis[5010];
int _back[5010];
void DFS(int now) {
  vis[now] = true;
  for (int i = np[now]; i != -1; i = nE[i].next) {
    int v = nE[i].v;
    if (vis[v] == false) {
      DFS(v);
    }
  }
}
int main() {
  cin >> n >> m >> s;
  init();
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    insert(u, v);
  }
  memset(dfn, 0, sizeof(dfn));
  times = cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (dfn[i] == 0) {
      dfs(i);
    }
  }
  ninit();
  for (int u = 1; u <= n; u++) {
    for (int i = p[u]; i != -1; i = E[i].next) {
      int v = E[i].v;
      if (belong[u] != belong[v]) {
        ninsert(belong[u], belong[v]);
        _back[belong[v]]++;
      }
    }
  }
  memset(vis, false, sizeof(vis));
  int ans = 0;
  DFS(belong[s]);
  for (int i = 1; i <= cnt; i++) {
    if (vis[i] == false && _back[i] == 0) {
      ans++;
      DFS(i);
    }
  }
  for (int i = 1; i <= cnt; i++) {
    if (vis[i] == false) {
      ans++;
      DFS(i);
    }
  }
  cout << ans << endl;
  return 0;
}
