Dr. X is a biologist, who likes rabbits very much and can do everything for them. 2012 is coming, and Dr. X wants to take some rabbits to Noah‘s Ark, or there are no rabbits any more.

A rabbit‘s genes can be expressed as a string whose length is l (1 ≤ l ≤ 100) containing only ‘A‘, ‘G‘, ‘T‘, ‘C‘. There is no doubt that Dr. X had a in-depth research on the rabbits‘ genes. He found that if a rabbit gene contained a particular gene segment, we could consider it as a good rabbit, or sometimes a bad rabbit. And we use a value W to measure this index.

We can make a example, if a rabbit has gene segment "ATG", its W would plus 4; and if has gene segment "TGC", its W plus -3. So if a rabbit‘s gene string is "ATGC", its W is 1 due to ATGC contains both "ATG"(+4) and "TGC"(-3). And if another rabbit‘s gene string is "ATGATG", its W is 4 due to one gene segment can be calculate only once.

Because there are enough rabbits on Earth before 2012, so we can assume we can get any genes with different structure. Now Dr. X want to find a rabbit whose gene has highest W value. There are so many different genes with length l, and Dr. X is not good at programming, can you help him to figure out the W value of the best rabbit.

There are multiple test cases. For each case the first line is two integers n (1 ≤ n ≤ 10)，l (1 ≤ l ≤ 100), indicating the number of the particular gene segment and the length of rabbits‘ genes.

The next n lines each line contains a string DNAi and an integer wi (|wi| ≤ 100), indicating this gene segment and the value it can contribute to a rabbit‘s W.

For each test case, output an integer indicating the W value of the best rabbit. If we found this value is negative, you should output "No Rabbit after 2012!".

2 4
ATG 4
TGC -3

1 6
TGC 4

4 1
A -1
T -2
G -3
C -4

4
4
No Rabbit after 2012!

Hint
case 1：we can find a rabbit whose gene string is ATGG(4), or ATGA(4) etc.
case 2：we can find a rabbit whose gene string is TGCTGC(4), or TGCCCC(4) etc.
case 3：any gene string whose length is 1 has a negative W.
 

HONG, Qize

lcy | We have carefully selected several similar problems for you:4053 4056 4059 4052 4051

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define inf 1000000000
using namespace std;
int n,l,w;
char s[110];
int L,rt,next[1010][26],end[1010],fail[1010],c[1010];
int dp[2][1010][1<<10];
int newnode(){
    memset(next[L],0,sizeof next[L]);
    c[L]=0;
    end[L++]=0;
    return L-1;
}
void init(){
    L=0;
    rt=newnode();
}
void insert(char *s,int z,int x){
    int le=strlen(s),now=rt;
    for(int i=0;i<le;i++){
        int x=s[i]-'A';
        if(!next[now][x]) next[now][x]=newnode();
        now=next[now][x];
    }
    end[now]+=z;
    c[now]|=(1<<x);
}
void build(){
    queue<int> q;
    int x=rt;
    for(int i=0;i<26;i++){
        if(next[x][i]){
            fail[next[x][i]]=rt;
            q.push(next[x][i]);
        }
    }
    while(!q.empty()){
        x=q.front();
        q.pop();
        for(int i=0;i<26;i++){
            if(!next[x][i]){
                next[x][i]=next[fail[x]][i];
            }else{
                fail[next[x][i]]=next[fail[x]][i];
                q.push(next[x][i]);
            }
        }
    }
}
void read(){
    for(int i=0;i<n;i++){
        scanf("%s %d",s,&w);
        insert(s,w,i);
    }
}
int f[4]={0,2,6,19};
void solve(){
    build();
    int cur=0;
    for(int j=0;j<L;j++){
        for(int k=0;k<(1<<n);k++) dp[cur][j][k]=-inf;
    }
    dp[cur][0][0]=0;
    int ans=-inf;
    for(int i=1;i<=l;i++){
        for(int j=0;j<L;j++){
            for(int k=0;k<(1<<n);k++) dp[1-cur][j][k]=-inf;
        }
        for(int j=0;j<L;j++){
            for(int k=0;k<(1<<n);k++){
                if(dp[cur][j][k]==-inf) continue;
                for(int p=0;p<4;p++){
                    int z=f[p],nxt=next[j][z];
                    int C=0,tmp=0,x=nxt;
                    while(x){
                        C|=c[x];
                        if((k&c[x])==0) tmp+=end[x];
                        x=fail[x];
                    }
                    dp[1-cur][nxt][k|C]=max(dp[1-cur][nxt][k|C],dp[cur][j][k]+tmp);
                }
            }
        }
        cur=1-cur;
        if(i==l){
            for(int j=0;j<L;j++){
                for(int k=0;k<(1<<n);k++){
                    ans=max(ans,dp[cur][j][k]);
                }
            }
        }
    }
    if(ans<0) puts("No Rabbit after 2012!");
    else printf("%d\n",ans);
}
int main(){
    while(~scanf("%d%d",&n,&l)){
        init();
        read();
        solve();
    }
    return 0;
}