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;
const long long INF = 1e18, MOD = 1e9 + 7, MAXN = 105;
const long double EPS = 1e-9, PI = acos(-1);
int32_t main() {
  int64_t n;
  cin >> n;
  vector<string> str(n);
  for (int64_t i = 0; i < n; ++i) cin >> str[i];
  vector<int64_t> next(26, -1), prev(26, -1);
  vector<bool> used(26);
  for (int64_t i = 0; i < n; ++i) {
    for (int64_t j = 0; j < str[i].size() - 1; ++j) {
      if ((next[str[i][j] - 'a'] != -1 &&
           next[str[i][j] - 'a'] != str[i][j + 1] - 'a') ||
          (prev[str[i][j + 1] - 'a'] != -1 &&
           prev[str[i][j + 1] - 'a'] != str[i][j] - 'a')) {
        cout << "NO" << endl;
        return 0;
      }
      used[str[i][j] - 'a'] = true;
      next[str[i][j] - 'a'] = str[i][j + 1] - 'a';
      prev[str[i][j + 1] - 'a'] = str[i][j] - 'a';
    }
    used[str[i].back() - 'a'] = true;
  }
  string answer = "";
  for (int64_t i = 0; i < 26; ++i) {
    if (used[i] && prev[i] == -1) {
      int64_t now = i;
      while (next[now] != -1) {
        answer += (char)(now + 'a');
        used[now] = false;
        now = next[now];
      }
      answer += (char)(now + 'a');
      used[now] = false;
    }
  }
  for (int64_t i = 0; i < 26; ++i) {
    if (used[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << answer << endl;
  return 0;
}
