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".
import sys      
import bisect
import math as mt

#input=sys.stdin.readline
#t=int(input())
t=1
mod=10**9+7
for _ in range(t):
    n=int(input())
    #n,k=map(int,input().split())
    #l=list(map(int,input().split()))
    d={}
    l1=[]
    for i in range(n):
        s=input()
        if d.get(len(s),-1)==-1:
            d[len(s)]=[]
        d[len(s)].append(s)    
        l1.append(len(s))
    l1=list(set(l1))
    l1.sort(reverse=True)
    n1=len(l1)
    ch=1
    for i in range(n1):
        s1=d[l1[i]][0]
        for  j in d[l1[i]]:
            #print(s1,j)
            if s1!=j:
                
                ch=0
                break
    #print(ch,d)
    for i in range (n1):
        s1=d[l1[i]][0]
        for j in range(i+1,n1):
            s2=d[l1[j]][0]
            #print(123,s1,s2)
            if s1.count(s2)==0:
                ch=0
                break
    if ch==0:
        print("NO")
    else:
        print("YES")
        for i in range(n1-1,-1,-1):
            for j in d[l1[i]]:
                print(j)
                
                
    
    