Description

Polycarp is working on a new operating system called BerOS. He asks you to help with implementation of a file suggestion feature.

There are n files on hard drive and their names are f 1 , f 2 , … , f n . Any file name contains between 1 and 8 characters, inclusive. All file names are unique.

The file suggestion feature handles queries, each represented by a string s . For each query s it should count number of files containing s as a substring (i.e. some continuous segment of characters in a file name equals s ) and suggest any such file name.

For example, if file names are “read.me”, “hosts”, “ops”, and “beros.18”, and the query is “os”, the number of matched files is 2 (two file names contain “os” as a substring) and suggested file name can be either “hosts” or “beros.18”.

Input

The first line of the input contains integer n ( 1 ≤ n ≤ 10000 ) — the total number of files.

The following n lines contain file names, one per line. The i -th line contains f i — the name of the i -th file. Each file name contains between 1 and 8 characters, inclusive. File names contain only lowercase Latin letters, digits and dot characters (’.’). Any sequence of valid characters can be a file name (for example, in BerOS “.”, “…” and “…” are valid file names). All file names are unique.

The following line contains integer q ( 1 ≤ q ≤ 50000 ) — the total number of queries.

The following q lines contain queries s 1 , s 2 , … , s q , one per line. Each s j has length between 1 and 8 characters, inclusive. It contains only lowercase Latin letters, digits and dot characters (’.’).

Output

Print q lines, one per query. The j -th line should contain the response on the j -th query — two values c j and t j , where

c j is the number of matched files for the j -th query, t j is the name of any file matched by the j -th query. If there is no such file, print a single character ‘-’ instead. If there are multiple matched files, print any. Sample Input

Input 4 test contests test. .test 6 ts . st. .test contes. st Output 1 contests 2 test. 1 test. 1 .test 0 - 4 .test

尋找子串，不能用KMP，因為已知串的長度都不超過8，所以就可以直接找已知串的所有子串存到map裡 定義三個map，一個記錄這個子串出現的次數，一個用來標記這個子串是否找過，另一個儲存這個子串的母串（任意一個） 程式碼：

#include<iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <map>
#include <cstdlib>
#include <algorithm>
#define ll long long
#define INF 0x3f3f3f
const int MAX=1e5+10;
using namespace std;
map <string,string> ans;
map <string,int> cnt;
map <string,bool> vis;
char a[20];
string str;
int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        scanf("%s",a);
         vis.clear();       
        int len=strlen(a);
        for(int i=0;i<len;i++)
        {
            str="";       //清空str
            for(int j=i;j<len;j++)
            {
                str+=a[j];          //從串的第一位開始往後找所有子串，再從第二位開始
                if(!vis[str])       //這個子串之前未出現過
                {
                    cnt[str]++;         //子串出現次數加1
                    ans[str]=a;         //記錄母串
                    vis[str]=true;     //表示這個子串已經存在
                }
            }
        }
    }
    int q;
    cin>>q;
    while(q--)
    {
        scanf("%s",a);
        if(cnt[a]==0)
            cout<<"0 -"<<endl;
        else
            cout<<cnt[a]<<" "<<ans[a]<<endl;
    }
  
    return 0;
}