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;
long long mod = 1e9 + 7;
long long modexp(long long x, long long n) {
  if (n == 0) return 1;
  if (n % 2 == 0) {
    long long y = modexp(x, n / 2) % mod;
    return (y * y) % mod;
  }
  return (x * modexp(x, n - 1) % mod) % mod;
}
long long powr(long long x, long long n) {
  long long ans = 1;
  for (int i = 1; i <= n; i++) ans = ans * x;
  return ans;
}
const int MAX = 5005;
vector<int> v[MAX];
bool visited[MAX];
void bfs(int i) {
  visited[i] = 1;
  queue<int> q;
  q.push(i);
  while (!q.empty()) {
    i = q.front();
    q.pop();
    for (int j = 0; j < v[i].size(); j++) {
      if (!visited[v[i][j]]) {
        visited[v[i][j]] = 1;
        q.push(v[i][j]);
      }
    }
  }
}
bool mark[MAX];
int track[MAX];
int cnt = 0;
int curr = 0;
void dfs(int i) {
  mark[i] = 1;
  track[curr++] = i;
  cnt++;
  for (int j = 0; j < v[i].size(); j++) {
    if (visited[v[i][j]] == 1) continue;
    if (mark[v[i][j]] == 0) {
      dfs(v[i][j]);
    }
  }
}
void dfs2(int i) {
  visited[i] = 1;
  for (int j = 0; j < v[i].size(); j++) {
    if (visited[v[i][j]] == 0) {
      dfs2(v[i][j]);
    }
  }
}
int main() {
  int n, m;
  scanf("%d", &n), scanf("%d", &m);
  int src;
  scanf("%d", &src);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d", &x), scanf("%d", &y);
    v[x].push_back(y);
  }
  bfs(src);
  int ans = 0;
  vector<pair<int, int> > vt;
  int idx = -1;
  int maxi = 0;
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) {
      curr = 0;
      cnt = 0;
      dfs(i);
      vt.push_back(make_pair(cnt, i));
      for (int j = 0; j < curr; j++) {
        mark[track[j]] = 0;
      }
    }
  }
  sort(vt.begin(), vt.end());
  reverse(vt.begin(), vt.end());
  for (int i = 0; i < vt.size(); i++) {
    if (!visited[vt[i].second]) {
      dfs2(vt[i].second);
      ans++;
    }
  }
  printf("%d\n", ans);
  return 0;
}
