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 push(s,l=0,r=-1):
    #print(l,r,a)
    if not a:
        a.insert(0,s)
        return(1)
    if r==-1:
        r=len(a)
    if l>=r:
        a.insert(r,s)
        return(1)
    m=(l+r)//2
    if len(s)<len(a[m]):
        if s not in a[m]:
            return(0)
        return(push(s,l,m))
    elif len(s)>len(a[m]):
        if a[m] not in s:
            return(0)
        return(push(s,m+1,r))
    elif s==a[m]:
        a.insert(m,s)
        return(1)
    else:
        return(0)
n=int(input())
a=[]
for i in range(n):
    s=input()
    if not push(s):
        print("NO")
        quit()
print("YES")
print('\n'.join(a))
