Problem: You are given n 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 n (1 ≤ n ≤ 10^{5}) — the number of words.

The i-th of the next n lines contains string s_{i} consisting lowercase alphabet letters — the i-th word. It is guaranteed that the sum of the total word length is equal or less than 10^{6}. Each word contains at least one vowel.

Output

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

In next 2m lines, print m 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.

Solution (in JAVA):

Solution:
import java.util.*;
import java.lang.*;
/*
  Now I am getting time limit on 5th test case
  probably because of remove operation
  maybe if I will use frequency array this problem will be solved))
  but if remove operation is costy then add operation is also costy, isn't it??  
  At least I implemented correct solution :)
  Now it is a matter of improving time cost from here and there :)
*/
public  class Main {
  public static class Pairs{
    String s1;
    String s2;
  }

  static Scanner in;  
  static final char[] vowels = {'a','e','i','o','u'};
  private static boolean isVowel(char ch) {
        return ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u';
    }

  public static int vowelCount(String s){
    int count=0;
    for(int i=0;i<s.length();i++){
      if(isVowel(s.charAt(i))){
          count++;
      }
    }
    return count;
  }

  public static int lastVowel(String s){
    int lastV=0;
    for(int i=s.length()-1;i>=0;i--){      
      if(isVowel(s.charAt(i))){
        for(int j=0; j<5; j++){
          if(s.charAt(i)==vowels[j]){
            lastV=j;
            return lastV;
          }          
        }
        
      }
    }
    return lastV;
  }
  static final int LEN = 1000001;

  public static void solve(){
    //System.out.println("here");
    //int[][] vowelFrequency = new int[LEN][6];    
    HashMap<Integer,HashMap<Integer,Queue<String>>> wordsByVowels = new HashMap<>();
    

    int n = in.nextInt();
    //String[] words = new String[n];
    for(int i=0; i<n;i++){
      String s = in.next();
      //words[i] = s;
      int vowelCount = vowelCount(s);
      int lastV = lastVowel(s);//a=1,e=2,i=3,o=4,u=5
      //vowelFrequency[vowelCount][lastV]++;
      //store index of a word
      if(!wordsByVowels.containsKey(vowelCount)){
        HashMap<Integer,Queue<String>> hm  =new HashMap<>();      
        wordsByVowels.put(vowelCount,hm);
      }
      if(!wordsByVowels.get(vowelCount).containsKey(lastV)){
        wordsByVowels.get(vowelCount).put(lastV,new LinkedList<String>());
      }
      //if(wordsByVowels.get(vowelCount))
      wordsByVowels.get(vowelCount).get(lastV).add(s);
    }

    //now separate as complete 
    Queue<Pairs> complete = new LinkedList<>();
    Queue<Pairs> incomplete = new LinkedList<>();
    for(Map.Entry<Integer , HashMap<Integer,Queue<String>> > entry: wordsByVowels.entrySet()){
      HashMap<Integer,Queue<String>> lastMap = entry.getValue();
      //iterate over Strings with same number of vowels
      for(Map.Entry<Integer,Queue<String>> e: lastMap.entrySet()){
            //take stack which contains elements with similar endVowel
            Queue<String> st = e.getValue();
            while(st.size()>=2){
              Pairs p = new Pairs();
              p.s1 = st.poll();              
              p.s2 = st.poll();              
              complete.add(p); 
            }
      }
      Queue<String> remainings = new LinkedList<>();

      for(Map.Entry<Integer,Queue<String>> e: lastMap.entrySet()){
            //take stack which contains elements with similar endVowel
            Queue<String> st = e.getValue();
            if(st.size()>0){
              remainings.add(st.poll());
            }
      }
      while(remainings.size()>=2){
        Pairs p = new Pairs();
        p.s1 = remainings.poll();
        p.s2 =  remainings.poll();
        incomplete.add(p);    
      }
    }
    
    //System.out.println("here");
//now we have both complete and incomplete and we can mix them well first mix completes with incompletes then mix completes if left any
  //System.out.printf("com %d inc %d\n",complete.size(),incomplete.size());
  //System.out.printf("com %d inc %d\n",complete.size(),incomplete.size());
  int lc = 0;
  if(complete.size()>incomplete.size()){
      lc=incomplete.size()+ (complete.size()-incomplete.size())/2;
  }else{
    lc = complete.size();
  }

  System.out.println(lc);
  

  while(complete.size()>0 && incomplete.size()>0){
    Pairs inc =incomplete.poll();
    Pairs com =complete.poll();
    System.out.printf("%s %s\n%s %s\n",inc.s1,com.s1,inc.s2,com.s2);
  }

  while(complete.size()>=2){
    Pairs com =complete.poll();
    Pairs com2 =complete.poll();
    System.out.printf("%s %s\n%s %s\n",com2.s1,com.s1,com2.s2,com.s2);
  }
    



  }


  public static void main(String[] args) {
    in = new Scanner(System.in);
    solve();
  }
}