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;
int main() {
  int n;
  cin >> n;
  vector<string> a(n);
  vector<int> g(26, -1), p(26, -1);
  vector<bool> st(26, false), vis(26, false);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  for (auto el : a) {
    for (int i = 0; i < el.size(); ++i) {
      int t = el[i] - 'a';
      vis[t] = true;
      if (i < el.size() - 1 && g[t] == -1) {
        g[t] = el[i + 1] - 'a';
        if (p[el[i + 1] - 'a'] != -1 && p[el[i + 1] - 'a'] != el[i] - 'a') {
          cout << "NO";
          return 0;
        } else {
          p[el[i + 1] - 'a'] = el[i] - 'a';
        }
      } else if (i < el.size() - 1 && g[t] == el[i + 1] - 'a' ||
                 i == el.size() - 1) {
        continue;
      } else {
        cout << "NO";
        return 0;
      }
    }
  }
  for (int i = 0; i < 26; ++i) {
    for (int j = g[i]; j != -1; j = g[j]) {
      if (j == i) {
        cout << "NO";
        return 0;
      }
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (vis[i] && p[i] == -1) {
      for (int j = i; j != -1; j = g[j]) {
        cout << char('a' + j);
      }
    }
  }
}
