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;
set<int> E[26];
bool V[26];
int indeg[26];
string ans;
void dfs(int i) {
  V[i] = 1;
  if (ans != "NO") ans += 'a' + i;
  if (E[i].size() > 1) ans = "NO";
  for (auto x : E[i]) {
    if (!V[x])
      dfs(x);
    else
      ans = "NO";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  string str;
  for (auto &x : V) x = 1;
  for (int i = 0; i < n; ++i) {
    cin >> str;
    for (int j = 0; j < str.size(); ++j) {
      V[str[j] - 'a'] = 0;
      if (j > 0)
        E[str[j - 1] - 'a'].insert(str[j] - 'a'), indeg[str[j] - 'a']++;
    }
  }
  for (int i = 0; i < 26; ++i)
    if (!V[i] and indeg[i] == 0) dfs(i);
  for (int i = 0; i < 26; ++i)
    if (!V[i]) ans = "NO";
  cout << (ans.empty() ? "NO" : ans) << endl;
  return 0;
}
