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;
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < (int)(t.size()); i++) s << (i ? " " : "") << t[i];
  return s;
}
template <typename T>
istream& operator>>(istream& s, vector<T>& t) {
  for (int i = 0; i < (int)(t.size()); i++) s >> t[i];
  return s;
}
template <typename T, typename U>
ostream& operator<<(ostream& s, pair<T, U> t) {
  s << "(" << t.first << "," << t.second << ")";
  return s;
}
template <typename T, typename U>
istream& operator>>(istream& s, pair<T, U>& t) {
  s >> t.first >> t.second;
  return s;
}
const int N = 26;
int after[N], before[N];
bool seen[N];
bool needed[N];
bool addConsecutive(int x, int y) {
  if (x == y) return false;
  if (after[x] == y) return true;
  if (after[x] != -1) return false;
  if (before[y] != -1) return false;
  after[x] = y;
  before[y] = x;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  memset(after, -1, sizeof(after));
  memset(before, -1, sizeof(before));
  int n;
  cin >> n;
  for (int i = 0; i < (int)(n); i++) {
    string s;
    cin >> s;
    for (int j = 0; j < (int)(s.size()); j++) needed[s[j] - 'a'] = true;
    for (int j = 0; j < (int)(s.size() - 1); j++) {
      int x = s[j] - 'a';
      int y = s[j + 1] - 'a';
      bool ok = addConsecutive(x, y);
      if (!ok) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  vector<string> parts;
  for (int i = 0; i < (int)(26); i++) {
    if (seen[i]) continue;
    if (!needed[i]) continue;
    string cs;
    int cur = i;
    while (before[cur] != -1) {
      cur = before[cur];
      if (cur == i) {
        cout << "NO" << endl;
        return 0;
      }
    }
    while (true) {
      seen[cur] = true;
      cs += (char)(cur + 'a');
      if (after[cur] == -1) break;
      cur = after[cur];
    }
    parts.push_back(cs);
  }
  sort(parts.begin(), parts.end());
  for (int i = 0; i < (int)(parts.size()); i++) cout << parts[i];
  cout << endl;
}
