You are given n strings. Each string consists of lowercase English letters. Rearrange (reorder) the given strings in such a way that for every string, all strings that are placed before it are its substrings.

String a is a substring of string b if it is possible to choose several consecutive letters in b in such a way that they form a. For example, string "for" is contained as a substring in strings "codeforces", "for" and "therefore", but is not contained as a substring in strings "four", "fofo" and "rof".

Input

The first line contains an integer n (1 ≤ n ≤ 100) — the number of strings.

The next n lines contain the given strings. The number of letters in each string is from 1 to 100, inclusive. Each string consists of lowercase English letters.

Some strings might be equal.

Output

If it is impossible to reorder n given strings in required order, print "NO" (without quotes).

Otherwise print "YES" (without quotes) and n given strings in required order.

Examples

Input

5
a
aba
abacaba
ba
aba


Output

YES
a
ba
aba
aba
abacaba


Input

5
a
abacaba
ba
aba
abab


Output

NO


Input

3
qwerty
qwerty
qwerty


Output

YES
qwerty
qwerty
qwerty

Note

In the second example you cannot reorder the strings because the string "abab" is not a substring of the string "abacaba".
//'main' method must be in a class 'Rextester'.
//Compiler version 1.8.0_111

import java.util.*;
import java.lang.*;

public class Rextester
{  
   void sort(String []s, int n)  
  { 
    for (int i=1 ;i<n; i++) 
    { 
        String temp = s[i]; 
        int j = i - 1; 
        while (j >= 0 && temp.length() < s[j].length()) 
        { 
            s[j+1] = s[j]; 
            j--; 
        } 
        s[j+1] = temp; 
    } 
 } 
    public static void main(String args[])
    {
      Rextester r =  new Rextester();
      Scanner s = new Scanner(System.in);
      int c = s.nextInt ();
      String[] st = new String[c];
      int n=0,Ch=0;
      while(n<=(c-1)){
      st[n]=s.next();
      n++;}
      r.sort(st , c);
     int i=0;
      while(i<=(c-1)){
      if((i>0) && Ch==0){
       if(st[i].contains(st[i-1]))
         Ch=0;
       else{
         Ch=1;
         break;}}
         i++;}
        if(Ch==1)
         System.out.println("NO");
        else{ System.out.println("YES");
              for(i=0;i<c;i++)
               System.out.println(st[i]+" ");
             }
    }
}