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".
import java.util.*;
public class Fest {
    public static void main(String[] args) {
        
       Scanner in = new Scanner(System.in);
       
       int n = in.nextInt();
       String s[] = new String[n];
       for(int i=0; i<n; i++)
           s[i] = in.next();
       Arrays.sort(s, new Comparator<String>()
       {
            public int compare(String a, String b)
            {
                return a.length()-b.length();
            }
        });
       boolean flag = true;
       for(int i=0; i<n-1; i++)
       {
           if(!kmp(s[i], s[i+1]))
           {
               flag = false;
               break;
           }
       }
       if(flag)
       {
           System.out.println("YES");
           for(int i=0; i<n; i++)
               System.out.println(s[i]);
       }
       else
            System.out.println("NO");
    }
    public static boolean kmp(String pat, String text)
    {
        int a[] = new int[pat.length()];
        
        prefix(pat,a);
        int count =0 ;
        int j=0;
        for(int i=0;i<text.length();)
        {
            if( text.charAt(i) == pat.charAt(j) )
            {
                i++;
                j++;
            }
            if( j == pat.length())
            {
                count++;
                j = a[j-1];
            }
            else if(i < text.length() && text.charAt(i)!=pat.charAt(j))
            {
                if(j!=0)
                    j=a[j-1];
                else
                    i++;
            }
        }
        if(count>0)
            return true;
        return false;
     }
    static void prefix(String s,int a[])
    {
        int j=0;
        a[0] = 0;
        for(int i=1;i<s.length();)
        {
            if(s.charAt(i) == s.charAt(j))
            { 
                
                j++ ; 
                a[i] = j;
                i++ ;
            }
            else
            {
                if(j!=0)
                    j = a[j-1];
                else
                {
                    a[i] = 0;
                    i++;
                }
            }
        }
    }
}
