// napisane Monikou
// namakana dynamika 2^M* M * N *K

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<vector>

#define SIZE(s) ((int)((s).size()))
#define FOREACH(it,vec) for(typeof((vec).begin())it=(vec).begin();it!=(vec).end();++it)
#define REP(i,n) for(int i=0;i<(int)(n);++i)

#define MAXN 100
#define MAXM 8
#define MOD 420047

using namespace std;

int N,M,K;
long long P[2][MAXN*MAXM][(1<<MAXM)];

bool is_ok(long long x, int t) {   // kontrolujem popis x a som v kroku t (rozsah 0 az M-1)
  long long lo,hi;
  lo = x & ((1LL<<(t+1))-1);
  hi = x ^ lo;
  if (lo & (lo<<1)) return false;
  if (hi & (hi<<1)) return false;
  return true;
}

long long add(long long w, bool x) { // pridam k w hodnotu x (0/1) a w urezem na poslednych M bitov
  w = w&((1LL<<(M-1))-1);
  w = w<<1;
  if (x) w|=1;
  return w;
}

void print() {
  REP(i,2)
    REP(j,3)
      REP(k,3)
        cout << "P[" << i << "][" << j << "][" << k << "] = " << P[i][j][k] << endl;
}

int main(void) {
  memset(P,0,sizeof(P));
  scanf("%d",&N); scanf("%d",&M); scanf("%d",&K);

  if (N==0) { printf("0\n"); return 0; }

  for(long long i=0;i<(1LL<<M);++i) {
    if (i & (i<<1)) continue;
    int b=0;
    REP(j,M) if (i&(1LL<<j)) ++b;
    P[M%2][b][i] = 1;
  }

  for(int n=M;n<N*M;++n) {
    memset(P[(n+1)%2],0,sizeof(P[(n+1)%2]));
    for(long long i=0;i<(1LL<<M);++i) {
      REP(k,K+1) {
        REP(d,2) {
          if (P[n%2][k][i] == 0) continue;    // nic nove nepridam
          if (k+d > K)  continue; // prilis vela ciernych
          if (d==1 && (i&(1LL<<(M-1)))) continue; // ak pridavam ciernu a vedla nej je tiez cierna
          long long nove = add(i,(bool)d);
          if (!is_ok(nove,n%M))  continue;    
          P[(n+1)%2][k+d][nove]+=P[n%2][k][i];
          P[(n+1)%2][k+d][nove]%=MOD;
        }
      }
    }
  }

  long long sum = 0;
  for(long long i=0;i<(1LL<<M);++i) {
    sum+= P[(N*M)%2][K][i];
    sum%= MOD;
  }
  printf("%lld\n",sum);
  return 0;
}