/**
    daca avem diamantul full de aur inseamna:
    1 * ( 1 + 2 + 3 + 4 + ... + 20)
    2 * ( 1 + 2 + 3 + 4 + ... + 20)
    ...
    20* ( 1 + 2 + 3 + 4 + ... + 20)
    -------------------------------
     =  (1 + 2 + 3 + 4 + ... + 20)^2 = (20*21)^2/2  =
     = 21 *20 * 21 * 20 / 4 = 21^2*10*10 = 441*100 = 44100 suma maxima/minima
*/

#include <cstdio>

#define Nmax 25
#define Smax 3*sup_limit + 5
#define sup_limit 44100
#define down_limit -sup_limit
#define MOD 10000

int DP[2][Smax]; /// DP[dim][s] = nr de comfiguratii de arie dim si cu suma s
int N,M,K;

using namespace std;

int abso(int x){
    if(x < 0) return -x;
    return x;
}

int main()
{
    freopen("diamant.in","r",stdin);
    freopen("diamant.out","w",stdout);

    scanf("%d%d%d",&N,&M,&K);
    int lc = 1,la = 0;
    DP[0][sup_limit] = 1;
    for(int i = 1; i <= N; ++i)
        for(int j =  1; j <= M; ++j)
        {
            for(int s = 0; s <= 2*sup_limit; ++s)
            {
                DP[lc][s] = DP[la][s]; /// am pus un 0
                if(s >= i*j) /// avem de unde aduna
                    DP[lc][s] = DP[lc][s] + DP[la][s-i*j]; /// am pus un 1;
                DP[lc][s] = DP[lc][s] + DP[la][s+i*j]; /// am pus un -1
                while(DP[lc][s] > 10000)
                    DP[lc][s] -= 10000;
            }
            lc ^= 1;
            la ^= 1;
        }
    if(K > sup_limit || K < down_limit )printf("0\n");
    else printf("%d\n",DP[la][K + sup_limit]);

    return 0;
}