There are n boys and m girls attending a theatre club. To set a play "The Big Bang Theory", they need to choose a group containing exactly t actors containing no less than 4 boys and no less than one girl. How many ways are there to choose a group? Of course, the variants that only differ in the composition of the troupe are considered different.

Perform all calculations in the 64-bit type: long long for С/С++, int64 for Delphi and long for Java.

Input

The only line of the input data contains three integers n, m, t (4 ≤ n ≤ 30, 1 ≤ m ≤ 30, 5 ≤ t ≤ n + m).

Output

Find the required number of ways.

Please do not use the %lld specificator to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specificator.

Examples

Input

5 2 5


Output

10


Input

4 3 5


Output

3
#include <bits/stdc++.h>
using namespace std;
static const double EPS = 1e-10;
template <typename T, typename S>
vector<T>& operator<<(vector<T>& a, S b) {
  a.push_back(b);
  return a;
}
template <typename T>
void operator>>(vector<T>& a, int b) {
  while (b--)
    if (!a.empty()) a.pop_back();
}
bool isprime(int n) {
  if (n < 2) return false;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return false;
  return true;
}
long long b_pow(long long x, long long n) {
  return n ? b_pow(x * x, n / 2) * (n % 2 ? x : 1) : 1ll;
}
string itos(int n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
static const long long mod = (1ll << 62);
long long powmod(long long a, long long n, long long mod) {
  if (n == 0) return 1;
  if (n % 2 == 1) return a * powmod(a, n - 1, mod) % mod;
  long long c = powmod(a, n / 2, mod);
  return c * c % mod;
}
long long ncr[40][40];
long long npr(int n, int k) {
  long long ans = 1;
  for (int i = 0; i < k; i++) ans *= (n - i);
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m, t;
  cin >> n >> m >> t;
  ncr[0][0] = 1;
  for (int i = 1; i <= 30; i++) {
    for (int j = 0; j <= i; j++) {
      if (j) ncr[i][j] += ncr[i - 1][j - 1];
      ncr[i][j] += ncr[i - 1][j];
    }
  }
  long long ans = 0;
  for (int b = 4; b <= n; b++) {
    for (int g = 1; g <= m; g++) {
      if (b + g == t) ans += ncr[n][b] * ncr[m][g];
    }
  }
  cout << ans << endl;
}
