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 N = 111111;
int to[27], n, m, x, y, now, du[27], Ba[27];
char s[N];
bool way[27][27];
bool vis[27];
void print(char c) {
  putchar(c + now);
  vis[c] = 1;
  if (to[c]) print(to[c]);
}
int main(void) {
  cin >> n;
  now = 'a' - 1;
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s + 1);
    int l = strlen(s + 1);
    vis[s[l] - now] = 1;
    for (int j = 1; j < l; ++j) {
      vis[s[j] - now] = 1;
      if (to[s[j] - now] != s[j + 1] - now) {
        if (!to[s[j] - now]) {
          to[s[j] - now] = s[j + 1] - now;
          if (Ba[s[j + 1] - now]) {
            puts("NO");
            return 0;
          }
          Ba[s[j + 1] - now] = s[j] - now;
        } else {
          puts("NO");
          return 0;
        }
      }
    }
  }
  for (int i = 1; i <= 26; ++i) way[i][to[i]] = 1, ++du[to[i]];
  for (int k = 1; k <= 26; ++k) {
    for (int i = 1; i <= 26; ++i) {
      for (int j = 1; j <= 26; ++j) {
        if (way[i][k] && way[k][j]) way[i][j] = 1;
      }
    }
  }
  for (int i = 1; i <= 26; ++i) {
    if (way[i][i]) {
      puts("NO");
      return 0;
    }
  }
  for (int i = 1; i <= 26; ++i) {
    if (vis[i] && !du[i]) print(i);
  }
  puts("");
}
