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 __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
long long int bigmod(long long int a, long long int p, long long int m) {
  if (p == 1) {
    return a % m;
  }
  if (p == 0) {
    return 1;
  }
  if (p % 2 == 0) {
    long long int ans = bigmod(a, p / 2, m);
    ans = (ans % m) * (ans % m);
    return ans % m;
  } else {
    long long int ans = bigmod(a, (p - 1), m);
    ans = (a % m) * (ans % m);
    return ans % m;
  }
}
long long int dp[200][200];
bool flag[200][200];
long long int ncr(long long int n, long long int r) {
  if (n == r) {
    return 1;
  }
  if (r == 1) {
    return n;
  }
  if (dp[n][r] != -1) {
    return dp[n][r];
  }
  dp[n][r] = ncr(n - 1, r - 1) + ncr(n - 1, r);
  return dp[n][r];
}
int main() {
  for (int i = 0; i < 150; i++) {
    for (int j = 0; j < 150; j++) {
      dp[i][j] = -1;
    }
  }
  long long int n, m, t, ans = 0, x;
  cin >> n >> m >> t;
  x = max((long long int)4, t - m);
  while (x <= n && t - x >= 1) {
    ans += (ncr(n, x) * ncr(m, t - x));
    x++;
  }
  cout << ans << endl;
  return 0;
}
