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.io.*;
import java.util.*;

public class Main {

    static int nBoys;
    static int nGirls;
    static int t;
    static long num;
    static int nActors;
    static char actors[];
    static long dp[][][];

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        nBoys = sc.nextInt();
        nGirls = sc.nextInt();
        t = sc.nextInt();
        nActors = nBoys + nGirls;
        actors = new char[nActors];

        for (int i = 0; i < nActors; i++) {
            if (i < nBoys) {
                actors[i] = 'B';
            } else {
                actors[i] = 'G';
            }
        }
        
        dp = new long[nActors+1][31][31];
        for(int i=0;i<=nActors;i++){
            for(int j=0;j<31;j++){
                for(int k=0;k<31;k++){
                    dp[i][j][k]=-1;
                }
            }
        }
        long x = permutations(0, 0, 0, 0);
        System.out.println(x);

    }

    public static long permutations(int index, int boys, int girls, int r) {
        if(r == t){
            if(boys>3 && girls>0){
                return 1;
            }
        }
        if(index >nActors-1){
            return 0;
        }
        if(r>t){
            return 0;
        }
        if(dp[index][boys][girls] !=-1){
            return dp[index][boys][girls];
        }
        long way1 = permutations(index+1, boys, girls, r);
        long way2 = 0;
        if(actors[index]== 'B'){
            way2=permutations(index+1,boys+1,girls,r+1);
        }else{
            way2=permutations(index+1,boys,girls+1,r+1);
        }
        return dp[index][boys][girls] = way1 + way2 ;
    }

}
