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.*;
import java.math.*;
import java.io.*;

public class CF988B {
	public CF988B() {
		FS scan = new FS();
		int n = scan.nextInt();
		String[] s = new String[n];
		for(int i = 0 ; i < n ; i++)
			s[i] = scan.next();
		for(int i = 0 ; i < n ; i++) {
			for(int j = i + 1 ; j < n ; j++) {
				if(s[i].contains(s[j])) {
					String tmp = s[j];
					s[j] = s[i];
					s[i] = tmp;
				}
			}
		}
		boolean valid = true;
		for(int i = 0 ; i < n - 1 ; i++)
			if(!s[i + 1].contains(s[i]))
				valid = false;
		if(valid) {
			System.out.println("YES");
			for(int i = 0 ; i < n ; i++)
				System.out.println(s[i]);
		} else {
			System.out.println("NO");
		}
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
	}
	public static void main(String[] args) { new CF988B(); }
}
