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.Arrays;
import java.util.Comparator;
import java.util.Scanner;

	
public class lolqop {


	public static void main(String args[])
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		String g[]=new String[n];
		for(int i=0;i<n;i++)
		{
			g[i]=scan.next();
		}
		Arrays.sort(g,new Comparator<String>()
				{

					@Override
					public int compare(String a, String b) {if(a.length()>b.length()){
							return 1;
							
						}else if(b.length()>a.length()){
							return -1;
						}else
						{
							return 0;
						}
					}
			
				});
		
	for(int i=0;i<n-1;i++)
	{
		if(g[i].length()==g[i+1].length())
		{
			if(!(g[i].equals(g[i+1]))){
				System.out.println("NO");
				return;
			}
		}
		else{
			
			
			//try to find the g[i] in g[i+1]
			boolean contains=false;
			//System.out.println(g[i+1].length()-g[i].length());
			for(int j=0;j<g[i+1].length()-g[i].length()+1;j++)
			{
				
			//	System.out.println("po");
				if(g[i+1].substring(j,j+g[i].length()).equals(g[i]))
				{
					contains=true;
					break;
				}
			}
			if(!contains)
			{
				System.out.println("NO");
				return;
			}
			
		}
		
	}
	System.out.println("YES");
	for(int i=0;i<n;i++)
	{
		System.out.println(g[i]);
	}
	}
	
}
