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())
last = [-1 for i in range(26)]
left = [-1 for i in range(26)]
right = [-1 for i in range(26)]
for i in range(n):
    a = list(map(lambda x: ord(x) - ord('a'), input()))
    #print(a)
    for x in a:
        if last[x] == i:
            print('NO')
            exit(0)
        last[x] = i
    for j in range(len(a) - 1):
        x = a[j]
        y = a[j + 1]
        if left[y] != -1 and left[y] != x or\
                right[x] != -1 and right[x] != y:
                    print('NO')
                    exit(0)
        left[y] = x
        right[x] = y

sol = []
cnt = 0
for b in range(26):
    cnt += (last[b] > -1)
    if left[b] == -1 and last[b] > -1:
        sol.append(b)
        x = b
        while right[x] != -1:
            x = right[x]
            sol.append(x)

if len(sol) == cnt:
    for x in sol:
        print(chr(x + ord('a')), end='')
    print()
else:
    print('NO')
