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>
#pragma optimize("", off)
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma comment(linker, "/STACK:36777216")
const long long inf = 0x3f3f3f3f3f3f3f3LL;
const long long mod = 1e9 + 7;
using namespace std;
template <class T>
void smin(T& a, T val) {
  if (a > val) a = val;
}
template <class T>
void smax(T& a, T val) {
  if (a < val) a = val;
}
template <typename T>
inline std::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {
  bool first = true;
  os << "[";
  for (auto i : v) {
    if (!first) os << ", ";
    os << i;
    first = false;
  }
  return os << "]";
}
const long long N = 100;
long long NCR[N][N], n, m, t;
long long nCr(long long n, long long r) {
  if (r > n - r) r = n - r;
  long long ans = 1;
  long long i;
  for (i = 1; i <= r; i++) {
    ans *= n - r + i;
    ans /= i;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  memset(NCR, -1, sizeof NCR);
  cin >> n >> m >> t;
  long long ans = 0;
  for (long long boys = 4; boys <= n; ++boys) {
    long long girls = t - boys;
    if (girls < 1 or girls > m) continue;
    ans += nCr(n, boys) * nCr(m, girls);
  }
  cout << ans;
  return 0;
}
