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 double eps = 1e-9;
const long long MOD = 1073741824;
const long long N = 30 + 9;
const long long inf = 1 << 30;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <typename T>
T exp(T b, T p) {
  T x = 1;
  while (p) {
    if (p & 1) x = (x * b);
    b = (b * b);
    p = p >> 1;
  }
  return x;
}
inline long long ones(long long n) {
  long long res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long comb[N][N];
void buildPT(long long n) {
  for (long long i = comb[0][0] = 1; i <= n; ++i)
    for (long long j = comb[i][0] = 1; j <= i; ++j)
      comb[i][j] = (comb[i - 1][j] + comb[i - 1][j - 1]) % MOD;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(0);
  long long n, m, t, ans = 0;
  cin >> n >> m >> t;
  buildPT(35);
  for (long long i = (4); i <= (n); i++) {
    if (t - i >= 1 and m >= t - i) {
      ans += (comb[n][i] * comb[m][(t - i)]);
    }
  }
  cout << ans;
}
