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> a(26, -1);
  vector<int> b(26, -1);
  vector<int> cnt(26);
  for (int i = 0; i < n; i++) {
    vector<bool> kek(26);
    string s;
    cin >> s;
    for (auto &c : s) {
      c -= 'a';
      cnt[c]++;
      if (kek[c]) {
        cout << "NO\n";
        return 0;
      } else {
        kek[c] = true;
      }
    }
    for (int i = 0; i + 1 < (int)s.size(); i++) {
      if (a[s[i]] != -1 && a[s[i]] != s[i + 1]) {
        cout << "NO\n";
        return 0;
      } else {
        if (b[s[i + 1]] != -1 && b[s[i + 1]] != s[i]) {
          cout << "NO\n";
          return 0;
        }
        a[s[i]] = s[i + 1];
        b[s[i + 1]] = s[i];
      }
    }
  }
  vector<string> res;
  vector<bool> vis(26);
  for (int i = 0; i < 26; i++) {
    if (b[i] == -1 && cnt[i]) {
      string s = "";
      int x = i;
      while (x != -1) {
        vis[x] = 1;
        s += (x + 'a');
        x = a[x];
      }
      res.push_back(s);
    }
  }
  for (int i = 0; i < 26; i++) {
    if (!vis[i] && cnt[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  sort(res.begin(), res.end(), [](string a, string b) { return a < b; });
  for (auto c : res) cout << c;
  cout << '\n';
}
