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 urm[1000];
int prec[1000];
bool is_urm[1000];
bool folosit[1000];
string ans;
void get_ans(char c) {
  is_urm[c] = 1;
  ans.push_back(c);
  if (urm[c]) get_ans(urm[c]);
}
int main() {
  set<int> total;
  int n;
  cin >> n;
  while (n--) {
    string s;
    cin >> s;
    for (int i(0); i < s.size() - 1; i++) {
      if (urm[s[i]] && urm[s[i]] != s[i + 1]) {
        cout << "NO\n";
        return 0;
      }
      if (is_urm[s[i + 1]] && prec[s[i + 1]] != s[i]) {
        cout << "NO\n";
        return 0;
      }
      urm[s[i]] = s[i + 1];
      is_urm[s[i + 1]] = 1;
      prec[s[i + 1]] = s[i];
    }
    for (auto i : s) {
      folosit[i] = 1;
      total.insert(i);
    }
  }
  for (int i('a'); i <= 'z'; i++)
    if (folosit[i] && !is_urm[i]) get_ans(i);
  if (total.size() != ans.size()) {
    cout << "NO\n";
    return 0;
  }
  cout << ans;
  return 0;
}
