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".
# 988B - Substrings Sort
# http://codeforces.com/contest/988/problem/B

def aux_strings(a, b):
    for i in range(len(b) - len(a) + 1):
        if a == b[i:len(a)+i]:
            return -1

    return None

def cmp_strings(a, b):
    if a == b:
        return 0
    elif len(a) < len(b):
        return aux_strings(a, b)
    elif len(b) < len(a):
        cmp = aux_strings(b, a)

        if cmp is not None:
            return aux_strings(b, a) * -1

        return None
    else:
        return None


def quicksort(v, p, r):
    if p < r:
        q = particionar(v, p, r)
        if q is None:
            return q

        if quicksort(v, p, q-1) is None:
            return None
        if quicksort(v, q+1, r) is None:
            return None

    return True


def particionar(v, p, r):
    x = v[p]
    i = p
    j = p + 1

    while j <= r:
        cmp = cmp_strings(v[j], x)
        if cmp is None:
            return cmp
        if cmp == -1:
            i += 1
            trocar(v, i, j)
        j += 1

    trocar(v, p, i)

    return i


def trocar(v, n, m):
    temp = v[n]
    v[n] = v[m]
    v[m] = temp


n = int(input())
vetor = []
for i in range(n):
    vetor.append(input())

ordenou = quicksort(vetor, 0, len(vetor)-1)

if ordenou:
    print('YES')
    for e in vetor:
        print(e)
else:
    print('NO')