Solve in Python:
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 ≤ 10^5) — 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 10^5.


-----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.
n=int(input())
m=[]
sc=[]
for i in range(n):
    m.append(input())
    sc.append(set(m[i]))
    if len(sc[i])!=len(m[i]):
        print('NO')
        break
else:
    i=0
    pX=False
    while i<len(m):
        j=i+1
        p=False
        while j<len(m):
            #print(m)
            z=len(sc[i].intersection(sc[j]))
            #a=len(sc[i])
            #b=len(sc[j])
            if m[i] in m[j]:
                m[i]=m[j]
                sc[i]=sc[j]
                sc.pop(j)
                m.pop(j)
                p=True
                break
            elif m[j] in m[i]:
                sc.pop(j)
                m.pop(j)
                j-=1
            elif z>0:
                if m[i][-z:]==m[j][:z]:
                    m[i]+=m[j][z:]
                elif m[j][-z:]==m[i][:z]:
                    m[i]=m[j]+m[i][z:]
                else:
                    pX=True
                    break
                sc[i]=set(m[i])
                m.pop(j)
                sc.pop(j)
                j-=1
                p=True
            j+=1
        if not p:
            i+=1
        if pX:
            print('NO')
            break
    if not pX:
        print(''.join(sorted(m)))