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.
n = int(input())
u = []
s = []
ok = True
for i in range(n):
    u.append(input())
    s.append(set(u[i]))
    if len(s[i]) != len(u[i]):
        print('NO')
        ok = False
        break
if ok:
    i = 0
    ok = False
    while i < len(u):
        j = i + 1
        p = False
        while j < len(u):
            z = len(s[i].intersection(s[j]))
            if u[i] in u[j]:
                u[i] = u[j]
                s[i] = s[j]
                s.pop(j)
                u.pop(j)
                p = True
                break
            elif u[j] in u[i]:
                s.pop(j)
                u.pop(j)
                j -= 1
            elif z > 0:
                if u[i][-z:] == u[j][:z]:
                    u[i] += u[j][z:]
                elif u[j][-z:] == u[i][:z]:
                    u[i] = u[j] + u[i][z:]
                else:
                    ok = True
                    break
                s[i] = set(u[i])
                u.pop(j)
                s.pop(j)
                j -= 1
                p = True
            j += 1
        if not p:
            i += 1
        if ok:
            print('NO')
            break
    if not ok:
        u.sort()
        print(''.join(u))
        






                
