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() {
  int n;
  cin >> n;
  vector<int> next(26, -1), last(26, -1);
  vector<bool> is_used(26, false);
  for (int i = 0; i < n; i++) {
    vector<bool> is_s(26, false);
    string s;
    cin >> s;
    for (char c : s) {
      c -= 'a';
      if (is_s[c]) {
        cout << "NO";
        return 0;
      } else {
        is_s[c] = true;
        is_used[c] = true;
      }
    }
    for (int i = 0; i < s.size() - 1; i++) {
      char c = s[i] - 'a', l = s[i + 1] - 'a';
      if (next[c] != -1 && next[c] != l) {
        cout << "NO";
        return 0;
      } else {
        if (last[l] != -1 && last[l] != c) {
          cout << "NO";
          return 0;
        }
        next[c] = s[i + 1] - 'a';
        last[s[i + 1] - 'a'] = c;
      }
    }
  }
  vector<string> ans;
  vector<bool> is_vis(26, false);
  for (int i = 0; i < 26; i++)
    if (is_used[i] && last[i] == -1) {
      ans.push_back("");
      for (int j = i; j != -1; j = next[j]) {
        is_vis[j] = true;
        ans.rbegin()->push_back(j + 'a');
      }
    }
  for (int i = 0; i < 26; i++)
    if (is_used[i] && !is_vis[i]) {
      cout << "NO";
      return 0;
    }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
}
