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>
using namespace std;
const int MAX_N = 100000;
char s[MAX_N + 4];
int ps[26], qs[26];
bool used[26];
int main() {
  int n;
  scanf("%d", &n);
  memset(ps, -1, sizeof(ps));
  memset(qs, -1, sizeof(qs));
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    used[s[0] - 'a'] = true;
    for (int j = 1; s[j]; j++) {
      int c0 = s[j - 1] - 'a', c1 = s[j] - 'a';
      used[c1] = true;
      if ((qs[c0] >= 0 && qs[c0] != c1) || (ps[c1] >= 0 && ps[c1] != c0)) {
        puts("NO");
        return 0;
      }
      qs[c0] = c1, ps[c1] = c0;
    }
  }
  int l = 0;
  for (int i = 0; i < 26; i++)
    if (used[i]) l++;
  int k = 0;
  for (int st = 0; st < 26; st++)
    if (used[st] && ps[st] < 0) {
      for (int u = st; u >= 0; u = qs[u]) s[k++] = 'a' + u;
    }
  s[k] = '\0';
  if (k == l)
    puts(s);
  else
    puts("NO");
  return 0;
}
