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.ArrayList;
import java.util.Scanner;

public class ProbB
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in);

        int nS = input.nextInt();

        boolean added;

        ArrayList<String> stringsByLength = new ArrayList<String>();

        for(int i = 0; i < nS; i++)
        {
            String next = input.next();

            added = false;

            if(stringsByLength.size() == 0)
            {
                stringsByLength.add(next);
                added = true;
            }
            else
            {
                for (int a = 0; a < stringsByLength.size(); a++)
                {
                    if (next.length() <= stringsByLength.get(a).length() && added == false)
                    {
                        stringsByLength.add(a, next);
                        added = true;
                    }
                }
            }
            if(!added)
            {
                stringsByLength.add(next);
            }
        }

        boolean allASubstring = true;

        for(int i = stringsByLength.size()-1; i >= 1; i--)
        {
            if(!stringsByLength.get(i).contains(stringsByLength.get(i-1)))
            {
                allASubstring = false;
            }
        }
        if(allASubstring)
        {
            System.out.println("YES");

            for (int a = 0; a < stringsByLength.size(); a++)
            {
                System.out.println(stringsByLength.get(a));
            }
        }
        else
        {
            System.out.println("NO");
        }
    }
}