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.io.*;
import java.util.*;
public class ACM_Anyone {

    private static final Scanner LS = new Scanner(System.in);
    private static final BufferedReader LF = new BufferedReader(new InputStreamReader(System.in));
    private static final BufferedWriter EF = new BufferedWriter(new OutputStreamWriter(System.out));
    public static void main(String[] args) throws IOException {
        
        int t = LS.nextInt();
        String vet [] = new String[t];
        for (int i = 0 ; i < t ; i ++)
            vet [i] = LS.next();
        vet = bubbleSort(vet);
        boolean p = true;
        for (int i = 0 ; i < t ; i ++) {
            for (int j = 0 ; j < i ; j ++) 
                if (! vet [i].contains(vet [j])) {
                    p = false;
                    break;
                }
            if (!p)
                break;
        }
        if (!p)
            System.out.println("NO");
        else {
            System.out.println("YES");
            for (int i = 0 ; i < t ; i ++)
                System.out.println(vet [i]);
        }
    }
    
    public static String[] bubbleSort(String[] vetor ) {
	int tamanho = vetor.length;
	for (int i = 0 ; i < tamanho - 1 ; i++) {
            for (int j = 0; j < tamanho - 1 - i; j++) {
                if(vetor[j].length() > vetor[ j+1 ].length() ) { 
                    String auxiliar = vetor[j];
                    vetor[j] = vetor[j + 1];
                    vetor[j + 1] = auxiliar;
		}
            }
	}
        return vetor;
    }
}
