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".
n=int(input())
l=[]
q=[]
w=[]
S=[]
D=[]
G=0;
C=1;
O=0;
for i in range(0,n):
    s=input().rstrip()
    l.append(len(list(s)))
    q.append(s)
while(1):
    if len(l)==0:
        O=1;
        break;
    t=min(l)
    f=l.index(t)
    A=list(q[f])
    del(l[f])
    del(q[f])
    if G==0:
        G=1;
        S.append(A)
    else:
        B=S[len(S)-1]
        H=0;
        for i in range(0,len(A)-len(B)+1):
            t=A[i:i+len(B)]
            if ''.join(t)==''.join(B):
                H=1;
                break;
        if H==1:
            S.append(A)
        else:
            C=0;
            break;
if C==0:
    print("NO")
else:
    print("YES")
    for i in range(0,len(S)):
        print(''.join(S[i]))
