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 anterior[30], proximo[30], need[30], vis[30];
int main() {
  memset(proximo, -1, sizeof proximo);
  memset(anterior, -1, sizeof anterior);
  int n;
  cin >> n;
  int bad = 0;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (char c : s) need[c - 'a'] = 1;
    for (int j = 0; j + 1 < s.size(); j++) {
      int cur = s[j] - 'a';
      int nxt = s[j + 1] - 'a';
      if (proximo[cur] != -1 && proximo[cur] != nxt) bad = 1;
      if (anterior[nxt] != -1 && anterior[nxt] != cur) bad = 1;
      proximo[cur] = nxt;
      anterior[nxt] = cur;
    }
  }
  if (bad) {
    cout << "NO" << endl;
    return 0;
  }
  string res = "";
  for (int i = 0; i < 26; i++)
    if (need[i] && anterior[i] == -1) {
      for (int j = i; j != -1; j = proximo[j]) {
        need[j] = 0;
        res += char(j + 'a');
      }
    }
  for (int i = 0; i < 26; i++)
    if (need[i]) {
      cout << "NO" << endl;
      return 0;
    }
  cout << res << endl;
  return 0;
}
