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;
const long long dx[4] = {-1, 0, 1, 0};
const long long dy[4] = {0, -1, 0, 1};
const long long dxx[] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long dyy[] = {0, 1, 1, 1, 0, -1, -1, -1};
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long modexpo(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modexpo((x * x) % M, n / 2, M);
  else
    return (x * modexpo((x * x) % M, (n - 1) / 2, M)) % M;
}
long long bexpo(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return bexpo(x * x, n / 2);
  else
    return x * bexpo(x * x, (n - 1) / 2);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
const long long mod = 1e9 + 7;
const long long N = 4e5 + 5;
const long long INF = 1e18;
long long C(long long a, long long b) {
  long long z = 1;
  for (long long i = b + 1; i <= a; i++) {
    z *= i;
    z = z / (i - b);
  }
  return z;
}
void solve() {
  long long n, m, t;
  cin >> n >> m >> t;
  ;
  long long ans = 0;
  for (long long i = 4; i <= n && i <= t - 1; i++) {
    if (m >= t - i) {
      ans += C(n, i) * C(m, t - i);
    }
  }
  cout << ans << "\n";
  ;
}
int32_t main() {
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
