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;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long indeg[26], to[26];
  bool needed[26], vis[26];
  for (long long i = 0; i < 26; i++) {
    indeg[i] = 0;
    to[i] = -1;
    needed[i] = vis[i] = false;
  }
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (long long j = 0; j < s.length(); j++) needed[s[j] - 'a'] = true;
    for (long long j = 0; j < s.length() - 1; j++) {
      indeg[s[j + 1] - 'a']++;
      if (to[s[j] - 'a'] == -1)
        to[s[j] - 'a'] = s[j + 1] - 'a';
      else if (to[s[j] - 'a'] != s[j + 1] - 'a') {
        cout << "NO\n";
        exit(0);
      }
    }
  }
  string res;
  for (long long i = 0; i < 26; i++) {
    if (vis[i] || indeg[i] > 0 || !needed[i]) continue;
    for (long long j = i; j != -1; j = to[j]) {
      if (vis[j]) {
        cout << "NO\n";
        exit(0);
      }
      res += (char)('a' + j);
      vis[j] = true;
    }
  }
  for (long long i = 0; i < 26; i++) {
    if (needed[i] && !vis[i]) {
      cout << "NO\n";
      exit(0);
    }
  }
  cout << res << '\n';
}
