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;
using ll = long long;
using ii = pair<int, int>;
using vb = vector<bool>;
using vi = vector<int>;
using vll = vector<ll>;
using vvi = vector<vi>;
using vvll = vector<vll>;
using vii = vector<ii>;
const ll MOD = 1000000007;
const int INF = 1000000000;
const double PI = 3.14159265358979323846;
void no() {
  cout << "NO\n";
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  vb occur(26, false), start(26, true);
  vi next(26, -1);
  int n;
  cin >> n;
  for (int i = 0, _i = (n); i < _i; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < static_cast<int>((s).size()); ++j) {
      occur[s[j] - 'a'] = true;
      if (j) {
        if (next[s[j - 1] - 'a'] != -1 && next[s[j - 1] - 'a'] != s[j] - 'a')
          no();
        next[s[j - 1] - 'a'] = s[j] - 'a';
        start[s[j] - 'a'] = false;
      }
    }
  }
  string ans;
  for (int i = 0, _i = (26); i < _i; ++i) {
    if (!occur[i] || !start[i]) continue;
    for (int j = i; j != -1; j = next[j]) {
      if (!occur[j]) no();
      ans.push_back(j + 'a');
      occur[j] = false;
    }
  }
  cout << (find((occur).begin(), (occur).end(), true) == occur.end() ? ans
                                                                     : "NO")
       << endl;
}
