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 main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  bitset<26> start;
  pair<int, int> letter[26];
  memset(letter, -1, sizeof(letter));
  int count[26];
  set<char> seen;
  string ans = "";
  string s;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s;
    start[s[0] - 'a'] = true;
    memset(count, 0, sizeof(count));
    for (int j = 0; j < s.size(); j++) {
      seen.insert(s[j]);
      if (count[s[j] - 'a'] > 0) goto bad;
      count[s[j] - 'a']++;
      if (j > 0) {
        if (letter[s[j] - 'a'].first != -1 &&
            letter[s[j] - 'a'].first != s[j - 1] - 'a')
          goto bad;
        if (letter[s[j - 1] - 'a'].second != -1 &&
            letter[s[j - 1] - 'a'].second != s[j] - 'a')
          goto bad;
        letter[s[j] - 'a'].first = s[j - 1] - 'a';
        letter[s[j - 1] - 'a'].second = s[j] - 'a';
      }
      if (j + 1 < s.size()) {
        if (letter[s[j] - 'a'].second != -1 &&
            letter[s[j] - 'a'].second != s[j + 1] - 'a')
          goto bad;
        if (letter[s[j + 1] - 'a'].first != -1 &&
            letter[s[j + 1] - 'a'].first != s[j] - 'a')
          goto bad;
        letter[s[j] - 'a'].second = s[j + 1] - 'a';
        letter[s[j + 1] - 'a'].first = s[j] - 'a';
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (start[i] && letter[i].first == -1) {
      for (int cur = i; cur != -1; cur = letter[cur].second) {
        ans.push_back(cur + 'a');
        seen.erase(cur + 'a');
      }
    }
  }
  if (seen.empty())
    cout << ans << '\n';
  else
    goto bad;
  return 0;
bad:
  cout << "NO" << '\n';
  return 0;
}
