This problem is to give m words and find out the type of words with at least q words and length of n. I n this problem, we first create a Trie for the word, then dp[i][j][k] on the tree represents the jth node of I on the tree, using the result of scheme K, which means that part 1 in the binary of K indicates that it has been used, Part 0 indicates that it has not been used, and finally add up the number of schemes with length N and single use > = P

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <map>
using namespace std;
#define LL long long
const int maxnode=1100;
const int sigma_size=27;
const int mod=20090717;
char s[1005];
int sz=1;

struct Node {
    int son[sigma_size];
    int val,fail;
}ch[maxnode];


struct AC {
    queue<int>Q;
    void init(int x) {ch[x].fail=ch[x].val=0;memset(ch[x].son,0,sizeof(ch[x].son));}
    int idx(char c) {return c-'a';}

    void insert(char s[],int v) {
        int u=0,n=strlen(s);
        for(int i=0;i<n;i++) {
            int c=idx(s[i]);
            if(!ch[u].son[c]) {
                init(sz);
                ch[u].son[c]=sz++;
            }
            u=ch[u].son[c];
        }
        ch[u].val=v;
    }

    void build() {
        for(int i=0;i<sigma_size;i++) if(ch[0].son[i]) Q.push(ch[0].son[i]);
        while(!Q.empty()) {
            int now=Q.front();Q.pop();
            int fail=ch[now].fail;
            for(int i=0;i<sigma_size;i++) {
                int nxt=ch[now].son[i];
                if(nxt) {
                    ch[nxt].fail=ch[fail].son[i];
                    Q.push(nxt);
                }
                else ch[now].son[i]=ch[fail].son[i];
                ch[ch[now].son[i]].val|=ch[ch[ch[now].fail].son[i]].val;
            }
        }
    }

}ans;

int num[1<<11+5];
LL dp[30][105][1025];
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);cout.tie(0);
    for(int i=1;i<(1<<10);i++) {
        for(int j=0;j<10;j++) {
            if(i&(1<<j)) num[i]++;
        }
    }
    int n,m,q;
    while(cin>>n>>m>>q){
        if(n==0&&m==0&&q==0) break;
        sz=1;
        ans.init(0);
        for(int i=0;i<m;i++) {
            cin>>s;
            ans.insert(s,(1<<i));
        }
        for(int i=0;i<=n;i++) {
            for(int j=0;j<sz;j++) {
                for(int k=0;k<=(1<<m);k++) {
                    dp[i][j][k]=0;
                }
            }
        }
        ans.build();
        dp[0][0][0]=1;
        for(int i=0;i<n;i++) {
            for(int j=0;j<sz;j++) {
                for(int k=0;k<(1<<m);k++) {
                    if(dp[i][j][k]>0) {
                        for(int x=0;x<26;x++) {
                            dp[i+1][ch[j].son[x]][k|ch[ch[j].son[x]].val]+=dp[i][j][k];
                            if(dp[i+1][ch[j].son[x]][(k|ch[ch[j].son[x]].val)]>=mod) dp[i+1][ch[j].son[x]][k|ch[ch[j].son[x]].val]-=mod;
                        }
                    }
                }
            }
        }
        LL ans=0;
        for(int i=0;i<(1<<m);i++) {
            if(num[i]>=q) {
                for(int j=0;j<sz;j++) {
                    ans=(ans+dp[n][j][i])%mod;
                }
            }
        }
        cout<<ans<<"\n";
    }
    return 0;
}