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;
int t = 1000001;
class myData {
 public:
  int t;
  int x;
  int y;
  int d;
  myData(int t, int x, int y, int d) : t(t), x(x), y(y), d(d) {}
  bool operator<(const myData &a) const { return (this->t < a.t); }
};
long long exponentMod(long long A, long long B, long long C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  long long y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y = (y * y) % C;
  } else {
    y = A % C;
    y = (y * exponentMod(A, B - 1, C) % C) % C;
  }
  return (long long)((y + C) % C);
}
pair<int, int> factorize(long long n) {
  int ans = 1;
  int c = 0;
  for (long long i = 2; i * i <= n; ++i) {
    while (n % i == 0) {
      if (i != 2) ans *= i, c++;
      n /= i;
    }
  }
  if (n != 1) {
    if (n != 2) ans *= n, c++;
  }
  return {ans, c};
}
map<long long, long long> bit;
int N1 = 2e5;
void update(long long i) {
  for (; i <= N1; i += i & (-i)) bit[i]++;
}
long long get(long long i) {
  long long ans = 0;
  for (; i > 0; i -= i & (-i)) ans += bit[i];
  return ans;
}
int root[200005];
long long si[200005];
int id(int x) {
  if (x == root[x]) return x;
  return root[x] = id(root[x]);
}
int now;
bool sortByVal(const myData &a, const myData &b) { return (a.t < b.t); }
long long ans1 = 0;
int spf[1000003];
long long calc[1000003 + 1];
long long nCk(long long n, long long k, long long mod) {
  return ((((calc[n] % mod) * exponentMod(calc[k], mod - 2, mod)) % mod) *
          exponentMod(calc[n - k], mod - 2, mod)) %
         mod;
}
void sieve() {
  spf[1] = 1;
  for (int i = 2; i < 1000003; i++) spf[i] = i;
  for (int i = 4; i < 1000003; i += 2) spf[i] = 2;
  for (int i = 3; i * i < 1000003; i++) {
    if (spf[i] == i) {
      for (int j = i * i; j < 1000003; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
set<int> getFactorization(int x) {
  set<int> ret;
  while (x != 1) {
    ret.insert(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
bool prime[1000003 + 1];
void sieve2() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= 1000003; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= 1000003; i += p) prime[i] = false;
    }
  }
}
int check_even(int x, int a[], int k, int n) {
  int c = 0;
  for (int i = 1; i < n; i++) {
    if (a[i] <= x) {
      c++;
      i++;
    }
  }
  if (c >= k)
    return 1;
  else
    return 0;
}
int check_odd(int x, int a[], int k, int n) {
  int c = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] <= x) {
      c++;
      i++;
    }
  }
  if (c >= k)
    return 1;
  else
    return 0;
}
vector<int> v[5050];
vector<bool> V(5050);
stack<int> ord;
void dfs1(int u) {
  V[u] = 1;
  for (auto x : v[u])
    if (!V[x]) dfs1(x);
}
void dfs2(int u) {
  V[u] = 1;
  for (auto x : v[u])
    if (!V[x]) dfs2(x);
  ord.push(u);
}
int main() {
  int n, m, a, b, c, ans = 0;
  cin >> n >> m >> a;
  for (int i = 0; i < m; ++i) cin >> b >> c, v[b].push_back(c);
  dfs1(a);
  for (int i = 1; i <= n; ++i)
    if (!V[i]) dfs2(i);
  V.assign(5050, 0);
  while (!ord.empty()) {
    int u = ord.top();
    ord.pop();
    if (!V[u]) dfs1(u), ++ans;
  }
  cout << ans;
}
