dp[l][r][k]/*表示区间[l,r]能否凑成k(W,I,N,G)字符*/
mp['W']=1;mp['I']=2;mp['N']=3;mp['G']=4;

#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

int na[],dp[][][],mp[],len;

char change[][][],toy[],h[]={' ','W','I','N','G'};

bool dfs(int l,int r,int k)

{

    if(l==r) return toy[l]==h[k];//len<=1,不能换了

    int& res=dp[l][r][k];

    if(~res) return res;//11,12行这样更快,不会T掉,可以看上面的记录

    for(int i=;i<=na[k];i++)

        for(int j=l;j<r;j++)

            if(dfs(l,j,mp[change[k][i][]])&&dfs(j+,r,mp[change[k][i][]]))

                return res=;

    return res=;

}

 int main()

 {

     memset(dp,-,sizeof(dp));

     for(int i=;i<=;i++) scanf("%d",&na[i]);

     for(int i=;i<=;i++)

     {

         for(int j=;j<=na[i];j++)

         {

             scanf("%s",&change[i][j]);

        }

    }

    scanf("%s",toy+);

    len=strlen(toy+);

    bool flag=;

    mp['W']=;mp['I']=;mp['N']=;mp['G']=;

    for(int i=;i<=;i++)

        if(dfs(,len,i))

        {

            flag=;

            printf("%c",h[i]);

        }

    if(!flag)    puts("The name is wrong!");

     return ;

 }