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;
int n, nxt[26], lst[26], qaq = 0;
bool chk[26] = {false};
char s[100005] = {'\0'}, outp[100] = {'\0'};
int main() {
  scanf("%d", &n);
  memset(nxt, -1, sizeof nxt);
  memset(lst, -1, sizeof lst);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    chk[s[0] - 'a'] = true;
    for (int j = 0; s[j + 1]; j++)
      if (nxt[s[j] - 'a'] != -1 && nxt[s[j] - 'a'] != s[j + 1] - 'a') {
        puts("NO");
        return 0;
      } else
        nxt[s[j] - 'a'] = s[j + 1] - 'a', lst[s[j + 1] - 'a'] = s[j] - 'a',
                   chk[s[j + 1] - 'a'] = true;
  }
  int st;
  for (int i = 0; i < 26; i++)
    if (lst[i] == -1 && chk[i]) {
      st = i;
      while (st != -1) {
        if (!chk[st]) {
          puts("NO");
          return 0;
        }
        outp[qaq++] = st + 'a';
        chk[st] = false;
        st = nxt[st];
      }
    }
  for (int i = 0; i < 26; i++)
    if (chk[i]) {
      puts("NO");
      return 0;
    }
  puts(outp);
  return 0;
}
