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 NO() {
  cout << "NO\n";
  exit(0);
}
int dis[26][26];
char buff[100];
bool pres[26];
bool get(int st) {
  memset(buff, 0, sizeof buff);
  buff[0] = st + 'a';
  for (int idx = 0; buff[idx]; idx++) {
    int i = buff[idx] - 'a';
    for (int j = 0; j < 26; j++) {
      int del = dis[i][j];
      if (del == 0) continue;
      if (idx + del < 0) return false;
      char& ch = buff[idx + del];
      if (ch == 0) ch = j + 'a';
      if (ch != j + 'a') return false;
    }
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int len = s.size();
    if (len > 26) NO();
    for (int j = 0; j < len; j++) {
      pres[s[j] - 'a'] = true;
      for (int k = 0; k < len; k++) {
        int a = s[j] - 'a', b = s[k] - 'a';
        if (dis[a][b] == 0) dis[a][b] = k - j;
        if (dis[a][b] != k - j) NO();
      }
    }
  }
  set<string> ans;
  for (int i = 0; i < 26; i++) {
    if (!pres[i]) continue;
    if (!get(i)) continue;
    for (char* ch = buff; *ch; ch++) {
      pres[*ch - 'a'] = 0;
    }
    ans.emplace(buff);
  }
  for (int i = 0; i < 26; i++)
    if (pres[i]) NO();
  if (ans.empty()) NO();
  for (auto& s : ans) cout << s;
  cout << '\n';
  return 0;
}
