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".
/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
// package div32;

import java.util.Scanner;

/**
 *
 * @author Mostafa
 */
public class Div32 {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        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() ;
        }
        for ( int i = 0 ; i < n - 1 ; ++i )
        {
            for ( int j = i + 1 ; j < n ; ++j )
            {
                if ( s[i].length() > s[j].length() )
                {
                    String r = s[i] ;
                    s[i] = s[j] ;
                    s[j] = r ;
                }
            }
        }
        boolean f = true ;
        for ( int i = 0 ; i < n - 1 ; ++i )
        {
            int g = 0 ;
            for ( int j = 0 ; j <= s[i+1].length() - s[i].length() ; ++j )
            {
                g = 0 ;
                int x = j ;
                while ( s[i].charAt(g) == s[i+1].charAt(x) )
                {
                    ++g ;
                    ++x ;
                    if ( g == s[i].length() )
                        break ;
                }
                if ( g == s[i].length() )
                    break ;
            }
            if ( g != s[i].length() )
            {
                f = false ;
                break ;
            }
        }
        if ( f )
        {
            System.out.println("YES");
            for ( int i = 0 ; i < n ; ++i )
                System.out.println(s[i]);
        }
        else
        {
            System.out.println("NO");
        }
    }   
}