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 abc {
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        String[] s = new String[n];
        int[] l = new int[n] ;
        String x = sc.nextLine();
        for(int i = 0; i < n; i++) {
            s[i] = sc.nextLine();
            l[i] = s[i].length();
        }
        /*for(int i = 0; i < n; i++) {
            System.out.println(s[i]);
        }*/
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n - i - 1; j++){
                if(l[j] > l[j+1] ) {
                    int f = l[j];
                    l[j] = l[j+1];
                    l[j+1] = f;
                    String t = s[j];
                    s[j] = s[j+1];
                    s[j+1] = t;
                }
            }
        }
        /*for(int i = 0; i < n; i++) {
                System.out.println(s[i]);
            }*/
        int c = 0;
        for(int i = 0; i < n-1; i++) {
            if(s[i+1].contains(s[i]) == true)
                c++;
        }
        
        if (c == n-1){
            System.out.println("YES");
            for(int i = 0; i < n; i++) {
                System.out.println(s[i]);
            }
        }
        else
            System.out.println("NO");
        
    }
}