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;
void optimise() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
long long int mod = 1e6 + 3;
void normalize(long long int &a) { a = (a + mod) % mod; }
long long int modmul(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a * b) % mod;
}
long long int modadd(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a + b) % mod;
}
long long int modsub(long long int a, long long int b) {
  a = a % mod;
  b = b % mod;
  normalize(a);
  normalize(b);
  return (a - b + mod) % mod;
}
long long int me(long long int x, long long int n) {
  x %= mod;
  if (n == 0) return 1;
  long long int u = me(x, n / 2) % mod;
  u = (u * u) % mod;
  if (n % 2) u = (u * x) % mod;
  return u;
}
long long int me1(long long int x, long long int n) {
  if (n == 0) return 1;
  long long int u = me1(x, n / 2);
  u = u * u;
  if (n % 2) u = u * x;
  return u;
}
inline long long int modInv(long long int a) { return me(a, mod - 2); }
inline long long int modDiv(long long int a, long long int b) {
  return modmul(a, modInv(b));
}
float power(float x, long long int y) {
  float temp;
  if (y == 0) return 1;
  temp = power(x, y / 2);
  if (y % 2 == 0)
    return temp * temp;
  else {
    if (y > 0)
      return x * temp * temp;
    else
      return (temp * temp) / x;
  }
}
long long int __gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return __gcd(b, a % b);
}
const long long int N = 5e3 + 1;
vector<long long int> vis(N);
vector<long long int> G[N];
vector<long long int> vis1(N);
void dfs(long long int v) {
  vis[v] = 1;
  for (auto y : G[v]) {
    if (vis[y] == 0) {
      dfs(y);
    }
  }
}
long long int cnt = 0;
void dfs1(long long int v) {
  vis1[v] = 1;
  for (auto y : G[v]) {
    if (vis1[y] == 0) {
      cnt++;
      dfs1(y);
    }
  }
}
bool compare(pair<long long int, long long int> x,
             pair<long long int, long long int> y) {
  return x.first > y.first;
}
void solve() {
  long long int n, m, c;
  cin >> n >> m >> c;
  for (long long int i = 0; i < m; i++) {
    long long int x, y;
    cin >> x >> y;
    G[x].push_back(y);
  }
  dfs(c);
  vector<pair<long long int, long long int> > v;
  for (long long int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      cnt = 0;
      dfs1(i);
      v.push_back({cnt, i});
      for (long long int i = 1; i <= n; i++) {
        vis1[i] = 0;
      }
    }
  }
  sort(v.begin(), v.end(), compare);
  long long int ans = 0;
  for (long long int i = 0; i < v.size(); i++) {
    if (vis[v[i].second] == 0) {
      ans++;
      dfs(v[i].second);
    }
  }
  cout << ans;
}
signed main() {
  optimise();
  long long int t;
  t = 1;
  cout << fixed << setprecision(5);
  while (t--) {
    solve();
    cout << "\n";
  }
}
