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-8;
const int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int inf = 1 << 30;
const long long Inf = 1LL << 62;
inline int Rand() { return (rand() << 16) | rand(); }
inline int sign(int x) { return (x > 0) - (x < 0); }
inline int comp(int x, int y) { return (x > y) - (x < y); }
inline int flp2(int x) {
  x |= x >> 1;
  x |= x >> 2;
  x |= x >> 4;
  x |= x >> 8;
  x |= x >> 16;
  return x - (x >> 1);
}
inline int clp2(int x) {
  x--;
  x |= x >> 1;
  x |= x >> 2;
  x |= x >> 4;
  x |= x >> 8;
  x |= x >> 16;
  return x + 1;
}
inline bool deg2(int x) { return (x & (x - 1)) == 0; }
inline int rev(int x) {
  x = ((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1);
  x = ((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2);
  x = ((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4);
  x = ((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8);
  x = ((x & 0xffff0000) >> 16) | ((x & 0x0000ffff) << 16);
}
inline bool eq(double a, double b) { return fabs(a - b) <= eps; }
int n, m, t, i, j, men, woman;
long long c[31][31], ans;
int main() {
  cin >> n >> m >> t;
  c[0][0] = 1;
  for (i = 1; i <= 30; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
  for (men = 4; men <= n; men++)
    if (1 <= t - men && t - men <= m) ans += c[n][men] * c[m][t - men];
  cout << ans;
  return 0;
}
