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, to[26], pr[26], us[26], ok = 1;
  cin >> n;
  for (int a = 0; a < 26; a++) {
    to[a] = -1, us[a] = 0, pr[a] = 0;
  }
  set<char> se;
  for (int a = 0; a < n; a++) {
    string s;
    cin >> s;
    int m = s.size();
    for (int a = 0; a < m - 1; a++) {
      int sy = s[a] - 'a', sy2 = s[a + 1] - 'a';
      if (to[sy] == -1 or to[sy] == sy2) {
        pr[sy2] = 1;
        to[s[a] - 'a'] = s[a + 1] - 'a';
      } else
        ok = 0;
      se.insert(s[a]);
    }
    se.insert(s[m - 1]);
  }
  string s;
  for (int a = 0; a < 26; a++) {
    if (se.find('a' + a) != se.end() and pr[a] == 0) {
      int v = a;
      while (v != -1) {
        if (us[v]) {
          ok = 0;
          break;
        } else {
          s.push_back('a' + v);
          us[v] = 1, v = to[v];
        }
      }
    }
  }
  if (ok and se.size() == s.size())
    cout << s;
  else
    cout << "NO";
}
