Farmer John has installed a new security system on the barn and now must issue a valid password to the cows in the herd. A valid password consists of L (3 <= L <= 15) different lower-case characters (from the traditional latin character set 'a'...'z'), has at least one vowel ('a', 'e', 'i', 'o', or 'u'), at least two consonants (non-vowels), and has characters that appear in alphabetical order (i.e., 'abc' is valid; 'bac' is not). 

Given a desired length L along with C lower-case characters, write a program to print all the valid passwords of length L that can be formed from those letters. The passwords must be printed in alphabetical order, one per line.

Input

* Line 1: Two space-separated integers, L and C 

* Line 2: C space-separated lower-case characters that are the set of characters from which to build the passwords

Output

* Lines 1..?: Each output line contains a word of length L characters (and no spaces). The output lines must appear in alphabetical order.

Sample Input

4 6
a t c i s w

Sample Output

acis
acit
aciw
acst
acsw
actw
aist
aisw
aitw
astw
cist
cisw
citw
istw




题目大概：

求字典序的全排列，但是要至少有1个元音字母和2和辅音字母。

思路：

dfs

代码：

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
int n,m;
char a[30];

int map[30];

int dfs(int k ,int t,int y,int f)
{
    if(t==n&&y>=1&&f>=2)
    {


                      for(int i=1;i<=m;i++)
        {
           if(map[i])cout<<a[i];
        }
        cout<<endl;


        return 0;
    }

    for(int i=k;i<=m;i++)
    {
     if(a[i]=='a'||a[i]=='e'||a[i]=='i'||a[i]=='o'||a[i]=='u')
     {
         map[i]=1;
         dfs(i+1,t+1,y+1,f);
         map[i]=0;
     }
     else
     {   map[i]=1;
         dfs(i+1,t+1,y,f+1);
         map[i]=0;

     }
    }


}
int main()
{
    cin>>n>>m;
    for(int i=1;i<=m;i++)
    {
        cin>>a[i];
    }
    sort(a+1,a+m+1);
    dfs(1,0,0,0);



    return 0;
}