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
import java.util.*;

public class Main {
  static long combs[][];

  public static void main (String[] args) {
    Scanner sc =  new Scanner(System.in);
		int n = sc.nextInt();
    int m = sc.nextInt();
    int t = sc.nextInt(); 
    long fac[] = new long[n+m+1];
    combs = new long[n+m+1][t+1];
    // for(long[] currComb : combs) {
    //   Arrays.fill(currComb, 0);
    // }
    for(int i = 1; i <= n; i++) {
      if(i == 1) {
        fac[i] = 1;
      }
      else {
        fac[i] = i * fac[i-1];
      }
    }
    long ans = 0;
    for(int i = 4; i <= n; i++) {
      int j = t - i;
      if(j < 1) {
        break;
      }
      else {
        if(j <= m) {
          ans += comb(n, i) * comb(m, j);
        }
      }
    }
    System.out.println(ans);
	}


  static long comb(int n, int k) {
    if(k == 0 || k == n) {
      return 1;
    }
    if(combs[n][k] == 0) {
      combs[n][k] = comb(n-1, k-1) + comb(n-1, k);
    }
    return combs[n][k];
  }
}

