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 maxn = 30;
int n;
string s, ans = "";
int mp[maxn], ct[maxn], in[maxn], out[maxn], used[maxn];
void span(int s) {
  ans = ans + char('a' + s);
  ct[s]--;
  if (mp[s] != -1) span(mp[s]);
}
int main() {
  cin >> n;
  memset(mp, -1, sizeof(mp));
  memset(ct, 0, sizeof(ct));
  for (int i = 0; i < n; ++i) {
    cin >> s;
    int len = s.size();
    if (len > 26) return 0 * printf("NO");
    for (int j = 0; j < len - 1; ++j) {
      ct[s[j] - 'a'] = 1;
      if (s[j] == s[j + 1]) return 0 * printf("NO");
      if (mp[s[j] - 'a'] == -1)
        mp[s[j] - 'a'] = s[j + 1] - 'a', in[s[j + 1] - 'a']++,
                  out[s[j] - 'a']++;
      else if (mp[s[j] - 'a'] != s[j + 1] - 'a')
        return 0 * printf("NO");
    }
    ct[s[len - 1] - 'a'] = 1;
  }
  for (int i = 0; i < 26; ++i)
    if (in[i] > 1 || out[i] > 1) return 0 * printf("NO");
  for (int s = 0; s < 26; ++s)
    if (ct[s] && !in[s]) span(s);
  int f = 0;
  for (int i = 0; i < 26; ++i)
    if (ct[i] != 0) f = 1;
  if (f) return 0 * printf("NO");
  cout << ans << endl;
  return 0;
}
