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 tempPrac
{

public static void main(String[] args)
{

    int n;
      Scanner sc = new Scanner(System.in);
      n = sc.nextInt();
      String[] s = new String[n];
      int[] sorting = new int[n];

      for(int i=0;i<n;i++){
          String temp = sc.next();
          sorting[i] = temp.length();
          s[i] = temp;
      }

      for(int i=0;i<n;i++){
          for(int j=i+1;j<n;j++){
              if(sorting[j]<sorting[i]){
                  int temp = sorting[j];
                  sorting[j] = sorting[i];
                  sorting[i] = temp;

                  String tempo = s[j];
                  s[j] = s[i];
                  s[i] = tempo;
              }
          }
      }
      if(n==1){
          System.out.println("YES");
          System.out.println(s[0]);
      }
      else{
        int f=1;
        for(int j=0;j<n-1;j++){
            if(!(s[j+1].contains(s[j]))){
                System.out.println("NO");
                f=0;
                break;
            }
            if(f==0) break;
        }
        if(f==1){
            System.out.println("YES");
            for(int  j=0;j<n;j++) {
                System.out.println(s[j]);
            }
        }
      }

}
}