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;
long long n, m, i, j, k, l, r, t, ans, x, y, fix[102], ii, z, xx;
int main() {
  cin >> n >> m >> t;
  for (i = 4; i < t; i++) {
    xx = i;
    y = t - i;
    if (xx > n || y > m) continue;
    for (ii = 0; ii < 102; ii++) fix[ii] = 0;
    for (z = 2; z <= n; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]++;
        }
      }
    }
    for (z = 2; z <= m; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]++;
        }
      }
    }
    for (z = 2; z <= n - xx; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]--;
        }
      }
    }
    for (z = 2; z <= xx; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]--;
        }
      }
    }
    for (z = 2; z <= y; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]--;
        }
      }
    }
    for (z = 2; z <= m - y; z++) {
      x = z;
      for (ii = 2; ii <= x; ii++) {
        while (x % ii == 0) {
          x /= ii;
          fix[ii]--;
        }
      }
    }
    k = 1;
    for (ii = 0; ii < 102; ii++) {
      for (j = 0; j < fix[ii]; j++) k *= ii;
    }
    ans += k;
  }
  cout << ans << endl;
  return 0;
}
