C. Beautiful Lyrics （CF vector pair的使用）

ou are given nn words, each of which consists of lowercase alphabet letters. Each word contains at least one vowel. You are going to choose some of the given words and make as many beautiful lyrics as possible.

Each lyric consists of two lines. Each line consists of two words separated by whitespace.

A lyric is beautiful if and only if it satisfies all conditions below.

• The number of vowels in the first word of the first line is the same as the number of vowels in the first word of the second line.
• The number of vowels in the second word of the first line is the same as the number of vowels in the second word of the second line.
• The last vowel of the first line is the same as the last vowel of the second line. Note that there may be consonants after the vowel.

Also, letters "a", "e", "o", "i", and "u" are vowels. Note that "y" is never vowel.

For example of a beautiful lyric,

"hello hellooowww"

"whatsup yowowowow"

is a beautiful lyric because there are two vowels each in "hello" and "whatsup", four vowels each in "hellooowww" and "yowowowow" (keep in mind that "y" is not a vowel), and the last vowel of each line is "o".

For example of a not beautiful lyric,

"hey man"

"iam mcdic"

is not a beautiful lyric because "hey" and "iam" don't have same number of vowels and the last vowels of two lines are different ("a" in the first and "i" in the second).

How many beautiful lyrics can you write from given words? Note that you cannot use a word more times than it is given to you. For example, if a word is given three times, you can use it at most three times.

Input

The first line contains single integer nn (1≤n≤1051≤n≤105) — the number of words.

The ii-th of the next nn lines contains string sisi consisting lowercase alphabet letters — the ii-th word. It is guaranteed that the sum of the total word length is equal or less than 106106. Each word contains at least one vowel.

Output

In the first line, print mm — the number of maximum possible beautiful lyrics.

In next 2m2m lines, print mm beautiful lyrics (two lines per lyric).

If there are multiple answers, print any.

Examples

input

14
wow
this
is
the
first
mcdics
codeforces
round
hooray
i
am
proud
about
that

output

3
about proud
hooray round
wow first
this is
i that
mcdics am

input

7
arsijo
suggested
the
idea
for
this
problem

output

0

input

4
same
same
same
differ

output

1
same differ
same same

Note

In the first example, those beautiful lyrics are one of the possible answers. Let's look at the first lyric on the sample output of the first example. "about proud hooray round" forms a beautiful lyric because "about" and "hooray" have same number of vowels, "proud" and "round" have same number of vowels, and both lines have same last vowel. On the other hand, you cannot form any beautiful lyric with the word "codeforces".

In the second example, you cannot form any beautiful lyric from given words.

In the third example, you can use the word "same" up to three times.

第一行第一个字中的元音数目与第二行第一个字中的元音数目相同。

第一行第二个字中的元音数目与第二行第二个字中的元音数目相同。

第一行的最后一个元音与第二行的最后一个元音相同。注意，元音后面可能有辅音。

#include<bits/stdc++.h>
#define pb push_back
#define mp make_pair//无需写出型别， 就可以生成一个pair对象
using namespace std;
const int N=1e6+10;
vector< pair<int,int> >ho,qi;
int d[130];
int vis[N][6];
string s[N];
int main()
{
    d['a']=1,d['e']=2,d['i']=3,d['o']=4,d['u']=5;
    int n;
    scanf("%d",&n);
    int kk=0;
    for(int i=1;i<=n;i++)
    {  //将具有相同数量元音字母的单词且最后元音字母相同的先配对
        cin>>s[i];
        int len=s[i].length(),tot=0,la;
        for(int j=0;j<len;j++)
           if(d[s[i][j]])
            tot++,la=d[s[i][j]];
        kk=max(tot,kk);
        if(vis[tot][la])
        {
            ho.pb(mp(vis[tot][la],i));
            vis[tot][la]=0;
        }
        else
            vis[tot][la]=i;
    }
    for(int i=1;i<=kk;i++)//边界为输入字符串最大元音数，
    {//将剩余的具有相同数量的元音字母的单词配对
        int la=0;
        for(int j=1;j<=5;j++)
        {
            if(vis[i][j])
            {
                if(la)
                {
                    qi.push_back(mp(vis[i][j],la));
                    la=0;
                }
                else
                    la=vis[i][j];
            }
        }
    }
    while(qi.size()<ho.size())//从ho中拿出一些放进qi里
    {
        qi.push_back(ho.back());
        ho.pop_back();//pop_back()函数删除最后一个元素
    }
    printf("%d\n",ho.size());
    for(int i=0;i<ho.size();i++)
    {
        cout<<s[qi[i].first]<<" "<<s[ho[i].first]<<endl<<s[qi[i].second]<<" "<<s[ho[i].second]<<endl;
    }
}