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 isSub(a, b):
    if len(a) > len(b):
        return False
    for i in range(0, len(b) - len(a) + 1):
        if a == b[i:i+len(a)]:
            return True
    return False

n = int(input().strip())

l = []
for i in range(n):
    l.append(input().strip())

max_len = 0
for e in l:
    if len(e) > max_len:
        max_len = len(e)

big_l = [[] for i in range(max_len + 1)]
for e in l:
    big_l[len(e)].append(e)

for small_l in big_l:
    if len(small_l) < 2:
        continue
    for i in range(1, len(small_l)):
        if small_l[i] != small_l[0]:
            print("NO")
            exit(0)

cache = None
for small_l in big_l:
    if len(small_l) == 0:
        continue
    if cache == None:
        cache = small_l[0]
        continue

    if not isSub(cache, small_l[0]):
        print("NO")
        exit(0)
    else:
        cache = small_l[0]

print("YES")
for small_l in big_l:
    for e in small_l:
        print(e)

