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.Scanner;

public class Main {

    private static void stringSort(String[] name) {
        for (byte i = 0; i < name.length - 1; i++) {
            for (byte j = (byte) (i + 1); j < name.length; j++) {
                if (name[i].length() > name[j].length()) {
                    String temp = name[i];
                    name[i] = name[j];
                    name[j] = temp;
                }
            }
        }
    }

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        byte n = in.nextByte(), max = 0;
        String s[] = new String[n];
        for (byte i = 0; i < n; i++) {
            s[i] = in.next();
        }
        stringSort(s);
        for (byte i = 0; i < n - 1; i++) {
            if (!s[i+1].contains(s[i])) {
                System.out.println("NO");
                return;
            }
        }
        System.out.println("YES");
        for (String str : s) {
            System.out.println(str);
        }
    }
}
