/**
*   Problema poate aprea imposibila, dar observam ca :
*   Suma maxima sete 44100 cand n = 20 si m = 20, deci daca x > 44100 rez = 0
*
*   Adunam sau scadem fiecare numar.
*
*   Time Complexity : O(N*M*S)
*
*   COROIAN DAVID, Satu Mare, ROMANIA
**/

#include <cstdio>

#define MOD 10000

using namespace std;

FILE *f, *g;

int dp[2][100009];

const int add = 45000;

int n, m, x;

void readFile()
{
    f = fopen("diamant.in", "r");

    fscanf(f, "%d%d%d", &n, &m, &x);

    fclose(f);
}

int rez;

void solve()
{
    int i, j;
    int s = 0;
    int k;
    int ant = 0, cr = 1;
    dp[ant][add] = 1;
    for(i = 1; i <= n; i ++)
    {
        for(j = 1; j <= m; j ++)
        {
            for(k = -s; k <= s; k ++)
            {
                dp[cr][k - i * j + add] += dp[ant][k + add];
                dp[cr][k         + add] += dp[ant][k + add];
                dp[cr][k + i * j + add] += dp[ant][k + add];

                dp[cr][k - i * j + add] %= MOD;
                dp[cr][k         + add] %= MOD;
                dp[cr][k + i * j + add] %= MOD;
            }

            s += i * j;

            for(k = -s; k <= s; k ++)
            {
                dp[ant][k + add] = dp[cr][k + add];
                dp[cr][k + add] = 0;
            }

        }
    }

    if(x < -s || x > s)
        rez = 0;

    else
        rez = dp[ant][x + add];
}

void printFile()
{
    g = fopen("diamant.out", "w");

    fprintf(g, "%d\n", rez);

    fclose(g);
}

int main()
{
    readFile();

    solve();

    printFile();

    return 0;
}