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".
def is_substring(string):
    try: dic_substrings[len(string)].append(string)
    except:
        dic_substrings[len(string)] = []
        dic_substrings[len(string)].append(string)

    for s in substrings:
        if len(s) <= len(string) and string.find(s) == -1: return False
        elif len(s) > len(string) and s.find(string) == -1: return False

    return True

n = int(raw_input())
dic_substrings = {}
substrings = []
no_substring = False

substring = raw_input()
dic_substrings[len(substring)] = []
dic_substrings[len(substring)].append(substring)
substrings.append(substring)

for i in xrange(n - 1):
    substring = raw_input()
    substrings.append(substring)

    if not no_substring:
        no_substring = not is_substring(substring)

if no_substring: print "NO"
else:
    print "YES"

    indexes = sorted(dic_substrings.keys())

    for i in indexes:
        strings = sorted(dic_substrings[i])

        print '\n'.join(strings)