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.
#include <bits/stdc++.h>
int next[26];
int occur[26];
int reached[26];
char str[100005];
std::string res;
int distinct = 0;
int main() {
  std::fill(next, next + 26, -1);
  int N;
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%s", str);
    for (int i = 0; str[i]; i++) {
      occur[str[i] - 'a'] = 1;
    }
    for (int i = 1; str[i]; i++) {
      if ((next[str[i - 1] - 'a'] != -1) &&
          (next[str[i - 1] - 'a'] != (str[i] - 'a'))) {
        printf("NO\n");
        return 0;
      }
      next[str[i - 1] - 'a'] = str[i] - 'a';
      reached[str[i] - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    distinct += occur[i];
  }
  for (int i = 0; i < 26; i++) {
    if (occur[i] && !reached[i]) {
      res.push_back('a' + i);
      distinct--;
      for (int k = next[i]; (k != i) && (k != -1); k = next[k]) {
        res.push_back('a' + k);
        distinct--;
        if (distinct < 0) {
          printf("NO\n");
          return 0;
        }
      }
    }
  }
  if (distinct > 0) {
    printf("NO\n");
    return 0;
  }
  printf("%s\n", res.c_str());
  return 0;
}
