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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const long long llmx = 9223372036854775807;
const int inf = 1e9;
struct kek {
  int x;
  int y;
  long long c;
};
bool isPrime(long long x) {
  if (x == 1) {
    return false;
  }
  for (long long i = 2; i <= (long long)sqrt(x); i++) {
    if (x % i == 0) {
      return false;
    }
  }
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long factorial(long long x) {
  long long sum = 1;
  for (long long i = 1; i <= x; i++) {
    sum *= i;
  }
  return sum;
}
int toint(string s) {
  int x = 0;
  for (int i = 0; i < (int)s.length(); i++) {
    x *= 10;
    x += (s[i] - '0');
  }
  return x;
}
string tostr(int x) {
  if (x == 0) {
    return "0";
  }
  string s = "";
  bool f = x < 0;
  x = abs(x);
  while (x != 0) {
    s += (x % 10) + '0';
    x /= 10;
  }
  string t = "";
  if (f) {
    t = "-";
  }
  for (int i = s.length() - 1; i >= 0; i--) {
    t += s[i];
  }
  return t;
}
long long bpow(long long a, long long b) {
  if (b == 0) {
    return 1;
  } else {
    if (b % 2 == 1) {
      return bpow(a, b - 1) * a;
    } else {
      long long r = bpow(a, b / 2);
      return r * r;
    }
  }
}
bool foo(int x) {
  while (x != 0) {
    int t = x % 10;
    if (t != 4 && t != 7) {
      return false;
    }
    x /= 10;
  }
  return true;
}
vector<int> g[5005];
int c[5005];
vector<int> b;
bool us[5005];
void dfs1(int x) {
  us[x] = true;
  c[x] = 0;
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (!us[y]) {
      dfs1(y);
    }
  }
}
void dfs2(int x) {
  us[x] = true;
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (!us[y]) {
      dfs2(y);
    }
  }
  b.push_back(x);
}
void dfs(int x, int cl) {
  us[x] = true;
  if (c[x] == -1) {
    c[x] = cl;
  }
  for (int i = 0; i < g[x].size(); i++) {
    int y = g[x][i];
    if (!us[y]) {
      dfs(y, cl);
    }
  }
}
int main() {
  int n, m, s;
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    c[i] = -1;
  }
  dfs1(s);
  for (int i = 1; i <= n; i++) {
    us[i] = false;
  }
  for (int i = 1; i <= n; i++) {
    if (!us[i]) {
      dfs2(i);
    }
  }
  int cl = 0;
  for (int i = 1; i <= n; i++) {
    us[i] = false;
  }
  reverse(b.begin(), b.end());
  for (int i = 0; i < b.size(); i++) {
    int x = b[i];
    if (!us[x] && c[x] == -1) {
      cl++;
      dfs(x, cl);
    }
  }
  cout << cl;
  return 0;
}
