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 __p(long long x) { cout << x; }
void __p(int32_t x) { cout << x; }
void __p(float x) { cout << x; }
void __p(double x) { cout << x; }
void __p(long double x) { cout << x; }
void __p(char x) { cout << '\'' << x << '\''; }
void __p(const char *x) { cout << '\"' << x << '\"'; }
void __p(const string &x) { cout << '\"' << x << '\"'; }
void __p(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __p(const pair<T, V> &x) {
  cout << '{';
  __p(x.first);
  cout << ',';
  __p(x.second);
  cout << '}';
}
template <typename T>
void __p(const T &x) {
  long long f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __p(i);
  cout << "}";
}
void _p() { cout << "]\n"; }
template <typename T, typename... V>
void _p(T t, V... v) {
  __p(t);
  if (sizeof...(v)) cout << ", ";
  _p(v...);
}
long long po(long long x, long long p, long long m = 1e18) {
  long long ans = 1;
  x %= m;
  while (p > 0) {
    if (p & 1) ans = (ans * x) % m;
    p = p >> 1;
    x = (x * x) % m;
  }
  return ans;
}
long long inv(long long x, long long m = 1000000007) {
  return po(x, m - 2, m) % m;
}
const long long N = 1e6 + 6;
long long n, m, k, x, y, z, a, b, c, d, i, j, l, r;
long long dp[31][31] = {};
void pre() {
  for (n = 0; n < 31; n++) {
    for (r = 0; 0 <= n ? r <= n : r >= n; 0 <= n ? r++ : r--) {
      if (r == n || r == 0)
        dp[n][r] = 1;
      else
        dp[n][r] = dp[n - 1][r] + dp[n - 1][r - 1];
    }
  }
}
void sol() {
  pre();
  cin >> n >> m >> x;
  c = 0;
  for (i = 4; 4 <= min(x - 1, n) ? i <= min(x - 1, n) : i >= min(x - 1, n);
       4 <= min(x - 1, n) ? i++ : i--) {
    if (m < x - i) continue;
    c += dp[n][i] * dp[m][x - i];
  }
  cout << c << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  sol();
  return 0;
}
