A substring of some string is called the most frequent, if the number of its occurrences is not less than number of occurrences of any other substring.

You are given a set of strings. A string (not necessarily from this set) is called good if all elements of the set are the most frequent substrings of this string. Restore the non-empty good string with minimum length. If several such strings exist, restore lexicographically minimum string. If there are no good strings, print "NO" (without quotes).

A substring of a string is a contiguous subsequence of letters in the string. For example, "ab", "c", "abc" are substrings of string "abc", while "ac" is not a substring of that string.

The number of occurrences of a substring in a string is the number of starting positions in the string where the substring occurs. These occurrences could overlap.

String a is lexicographically smaller than string b, if a is a prefix of b, or a has a smaller letter at the first position where a and b differ.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of strings in the set.

Each of the next n lines contains a non-empty string consisting of lowercase English letters. It is guaranteed that the strings are distinct.

The total length of the strings doesn't exceed 105.

Output

Print the non-empty good string with minimum length. If several good strings exist, print lexicographically minimum among them. Print "NO" (without quotes) if there are no good strings.

Examples

Input

4
mail
ai
lru
cf


Output

cfmailru


Input

3
kek
preceq
cheburek


Output

NO

Note

One can show that in the first sample only two good strings with minimum length exist: "cfmailru" and "mailrucf". The first string is lexicographically minimum.
import sys
n = int(input())
a=[]
for i in range(n):
	a.append([j for j in input()])

b=[]
for i in range(n):
	b.append({})
	for j in a[i]:
		if j in b[i]:
			print('NO')
			sys.exit()
		b[i][j]=1

c=[-1 for i in range(26)]
d=[-1 for i in range(26)]
e=[0 for i in range(26)]
for word in a:
	l=ord(word[0])-97
	e[l]=1
	for ch in word[1:]:
		ch1 = ord(ch)-97
		e[ch1]=1
		if c[l]==-1:
			c[l]=ch1
		elif c[l]!=ch1:
			print('NO')
			sys.exit()
		if d[ch1]==-1:
			d[ch1]=l
		elif d[ch1]!=l:
			print('NO')
			sys.exit()
		l=ch1

x=[]
def abc(c,d,e,i):
	e[i]=0
	if c[i]==-1:
		return
	else:
		i=c[i]
		x.append(chr(i+97))
		abc(c,d,e,i)
for i in range(26):
	if e[i]==1 and d[i]==-1:
		x.append(chr(i+97))
		abc(c,d,e,i)

if sum(e)>0:
	print('NO')
else:
	print(''.join(x))







# Made By Mostafa_Khaled