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 solve() {
  vector<int> next(26, -1);
  vector<int> prev(26, -1);
  vector<bool> isThere(26);
  int n;
  cin >> n;
  for (int _ = 0; _ < n; _++) {
    string s;
    cin >> s;
    set<char> v((s).begin(), (s).end());
    if ((int)v.size() != (int)s.size()) {
      cout << "NO";
      return;
    }
    for (auto i : v) isThere[i - 'a'] = 1;
    for (int i = 1; i < (int)s.size(); i++) {
      if (next[s[i - 1] - 'a'] == -1 and prev[s[i] - 'a'] == -1)
        next[s[i - 1] - 'a'] = s[i] - 'a', prev[s[i] - 'a'] = s[i - 1] - 'a';
      else if (next[s[i - 1] - 'a'] == -1) {
        if (prev[s[i] - 'a'] != s[i - 1] - 'a') {
          cout << "NO";
          return;
        }
        next[s[i - 1] - 'a'] = s[i] - 'a';
      } else if (prev[s[i] - 'a'] == -1) {
        if (next[s[i - 1] - 'a'] != s[i] - 'a') {
          cout << "NO";
          return;
        }
        prev[s[i] - 'a'] = s[i - 1] - 'a';
      } else {
        if (prev[s[i] - 'a'] != s[i - 1] - 'a') {
          cout << "NO";
          return;
        }
        if (next[s[i - 1] - 'a'] != s[i] - 'a') {
          cout << "NO";
          return;
        }
      }
    }
  }
  string s;
  for (int i = 0; i < 26; i++) {
    if (isThere[i] and prev[i] == -1) {
      int cur = i;
      while (cur != -1) {
        s.push_back(char(cur + 'a'));
        isThere[cur] = 0;
        cur = next[cur];
      }
    }
  }
  if (any_of((isThere).begin(), (isThere).end(), [](bool x) { return x; })) {
    cout << "NO";
    return;
  }
  cout << s;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
