#include <fstream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
ifstream fin ("cowfood.in");
ofstream fout("cowfood.out");
long long n,m,X,i,S,x,y,s,k,j,sol,D[10003][32],a[22][32],nr,M[32];
const long long MOD=3210121;
void backtrack(int pas)
{
    long long aux[32];
    if(pas==n+1)
    {
        if(!nr)
            return;
        s=0;
        for(int j=1;j<=k;j++)
            s+=M[j];
        if(s>S)
            return;
        if(nr&1)
            sol+=(D[S-s][k+1]%MOD);
        else
            sol-=(D[S-s][k+1]%MOD);
        if(sol<0)
            sol+=MOD;
        sol%=MOD;
    }
    else
    {
        for(int i=0;i<=1;i++)
        {
            if(i)
            {
                for(int j=1;j<=k;j++)
                {
                    aux[j]=M[j];
                    M[j]=max(M[j],a[pas][j]);
                }
                nr++;
            }
            backtrack(pas+1);
            if(i)
            {
                for(int j=1;j<=k;j++)
                    M[j]=aux[j];
                nr--;
            }
        }
    }
}
int main()
{
    fin>>k>>S>>n;
    for(i=1;i<=n;i++)
        for(j=1;j<=k;j++)
            fin>>a[i][j];
    for(j=0;j<=k+1;j++)
        D[0][j]=1;
    for(i=1;i<=S+1;i++)
        for(j=0;j<=k+1;j++)
            if(j<2)
                D[i][j]=1;
            else
                if(i==1)
                    D[i][j]=j;
                else
                    D[i][j]=(D[i-1][j]+D[i][j-1])%MOD;
    backtrack(1);
    sol=D[S][k+1]-sol;
    if(sol<0)
        sol+=MOD;
    ///scadem solutiile in care sunt toate valorile egale cu 0 (-1)
    ///si pe cele in care sunt k-1 valori egale cu 0. In acest caz se pun
    ///intre 1 si s bile in aceeasi cutie si de acea avem s*k
    sol-=(S*k+1);
    if(sol<0)
        sol+=MOD;
    fout<<sol;
    return 0;
}