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;
void fail() {
  cout << "NO" << endl;
  exit(0);
}
int main() {
  int n;
  cin >> n;
  vector<int> bef(26, -1), aft(26, -1);
  vector<bool> used(26, true);
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (auto& ch : s) ch -= 'a', used[ch] = false;
    for (int i = 0; i < s.size() - 1; ++i) {
      if (s[i + 1] == s[i]) fail();
      if (aft[s[i]] != -1 && aft[s[i]] != s[i + 1]) fail();
      aft[s[i]] = s[i + 1];
      if (bef[s[i + 1]] != -1 && bef[s[i + 1]] != s[i]) fail();
      bef[s[i + 1]] = s[i];
    }
  }
  vector<string> strs;
  for (int i = 0; i < 26; ++i) {
    if (used[i]) continue;
    int j = i;
    int cnt = 0;
    while (bef[j] != -1) {
      if (++cnt > 30) fail();
      j = bef[j];
    }
    string s1;
    while (j != -1) s1.push_back(j + 'a'), used[j] = true, j = aft[j];
    strs.push_back(s1);
  }
  sort(strs.begin(), strs.end());
  for (auto x : strs) cout << x;
  cout << endl;
}
