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;
class Solution {
 public:
  void solve(std::istream& in, std::ostream& out) {
    int n;
    in >> n;
    std::vector<int> prev(26, -1), next(26, -1);
    std::vector<bool> to_use(26);
    for (int i = 0; i < n; ++i) {
      std::string s;
      in >> s;
      for (int j = 0; j < s.length(); ++j) {
        if (j > 0) {
          if (prev[s[j] - 'a'] != -1 && prev[s[j] - 'a'] != s[j - 1] - 'a')
            return void(out << "NO\n");
          prev[s[j] - 'a'] = s[j - 1] - 'a';
        }
        if (j < s.length() - 1) {
          if (next[s[j] - 'a'] != -1 && next[s[j] - 'a'] != s[j + 1] - 'a')
            return void(out << "NO\n");
          next[s[j] - 'a'] = s[j + 1] - 'a';
        }
        to_use[s[j] - 'a'] = true;
      }
    }
    std::vector<std::deque<char>> ds;
    for (int i = 0; i < 26; ++i) {
      if (to_use[i]) {
        std::deque<char> d;
        d.push_back(i + 'a');
        to_use[i] = false;
        auto it = d.rbegin();
        int letter = i;
        while (it != d.rend() && letter != -1 && *it == letter + 'a') {
          ++it;
          letter = prev[letter];
        }
        if (it == d.rend()) {
          while (letter != -1) {
            if (!to_use[letter]) return void(out << "NO\n");
            d.push_front(letter + 'a');
            to_use[letter] = false;
            letter = prev[letter];
          }
        } else if (letter != -1 && *it != letter)
          return void(out << "NO\n");
        letter = next[i];
        while (letter != -1) {
          if (!to_use[letter]) return void(out << "NO\n");
          d.push_back(letter + 'a');
          to_use[letter] = false;
          letter = next[letter];
        }
        ds.push_back(std::move(d));
      }
    }
    std::sort(ds.begin(), ds.end());
    std::string s;
    for (auto& d : ds) s.insert(s.end(), d.begin(), d.end());
    out << s << '\n';
  }
};
void solve(std::istream& in, std::ostream& out) {
  out << std::setprecision(12);
  Solution solution;
  solution.solve(in, out);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  istream& in = cin;
  ostream& out = cout;
  solve(in, out);
  return 0;
}
