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;
vector<int> G[30];
int indeg[30], outdeg[30], edge[30][30], jest[30];
string res;
void Ans(int v) {
  res += (char)(v + 'a');
  jest[v] = 0;
  for (auto u : G[v]) Ans(u);
}
void no() {
  cout << "NO\n";
  exit(0);
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j + 1 < s.size(); j++) {
      int a = s[j] - 'a', b = s[j + 1] - 'a';
      if (!edge[a][b]) {
        jest[a] = jest[b] = 1;
        indeg[b]++, outdeg[a]++;
        edge[a][b] = 1;
        G[a].push_back(b);
      }
    }
    jest[s[0] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (indeg[i] > 1 || outdeg[i] > 1) no();
  for (int i = 0; i < 26; i++)
    if (!indeg[i] && jest[i]) Ans(i);
  for (int i = 0; i < 26; i++)
    if (jest[i]) no();
  cout << res << "\n";
}
