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 l[250];
int r[250];
int was[250];
stringstream ss;
void dfs(char c) {
  if (!c) return;
  if (was[c]) return;
  was[c] = 1;
  ss << c;
  dfs(r[c]);
}
int main() {
  set<char> ks;
  int n;
  cin >> n;
  for (int t = 0; t < n; ++t) {
    string k;
    cin >> k;
    memset(was, 0, sizeof was);
    for (int i = 0; i < k.size(); ++i) {
      ks.insert(k[i]);
      if (was[k[i]]) {
        cout << "NO" << endl;
        return 0;
      }
      was[k[i]] = 1;
      if (i + 1 == k.size()) continue;
      if (r[k[i]] != 0 && r[k[i]] != k[i + 1]) {
        cout << "NO" << endl;
        return 0;
      }
      if (l[k[i + 1]] != 0 && l[k[i + 1]] != k[i]) {
        cout << "NO" << endl;
        return 0;
      }
      r[k[i]] = k[i + 1];
      l[k[i + 1]] = k[i];
    }
  }
  memset(was, 0, sizeof was);
  for (auto i : ks) {
    if (l[i] == 0) dfs(i);
  }
  string resa;
  ss >> resa;
  if (resa.size() != ks.size()) {
    cout << "NO" << endl;
    return 0;
  }
  cout << resa << endl;
  return 0;
}
