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;
void Fastio() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long fact[1000006], inv[1000006];
long long fp(long long base, long long exp) {
  if (exp == 0) return 1;
  long long ans = fp(base, exp / 2);
  ans = (ans * ans) % 100000000000000000;
  if (exp % 2 != 0)
    ans = (ans * (base % 100000000000000000)) % 100000000000000000;
  return ans;
}
void calcFacAndInv(long long n) {
  fact[0] = inv[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = (i * fact[i - 1]);
    inv[i] = fp(fact[i], 100000000000000000 - 2);
  }
}
long long ncr(long long n, long long r) {
  return ((fact[n] * inv[r]) % 100000000000000000 * inv[n - r]) %
         100000000000000000;
}
long long npr(long long n, long long r) {
  return (fact[n] * inv[n - r]) % 100000000000000000;
}
long double NCR(int n, int m) {
  long double res = 1;
  m = min(m, n - m);
  for (int i = 0; i < m; i++) {
    res = res * (n - i) / (i + 1);
  }
  return res;
}
int main() {
  Fastio();
  calcFacAndInv(50);
  long long n, m, t, res = 0;
  cin >> n >> m >> t;
  for (int i = 4; i <= min(n, t - 1); i++) {
    if (m < t - i) continue;
    long long res1 = NCR(n, i), res2 = NCR(m, t - i);
    res += (res1 * res2);
  }
  cout << res << '\n';
  return 0;
}
