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 graph[27];
int grapht[27];
vector<bool> used(27, false);
vector<bool> opened(27, false);
string result;
void dfs(int node) {
  used[node] = true;
  opened[node] = true;
  if (graph[node] == -1) {
    result += (node + 'a');
    opened[node] = false;
    return;
  }
  if (opened[graph[node]]) {
    cout << "NO";
    exit(0);
  }
  dfs(graph[node]);
  opened[node] = false;
  result += (node + 'a');
}
void dfsn(int node) {
  opened[node] = true;
  if (graph[node] != -1) {
    if (opened[graph[node]]) {
      cout << "NO";
      exit(0);
    }
    dfsn(graph[node]);
  }
  opened[node] = false;
}
set<int> letter;
int main() {
  int n;
  cin >> n;
  fill(graph, graph + 27, -1);
  fill(grapht, grapht + 27, -1);
  while (n--) {
    string s;
    cin >> s;
    for (int i = 0; i < s.size() - 1; i++) {
      int b = s[i] - 'a';
      int e = s[i + 1] - 'a';
      if ((graph[b] != -1 && graph[b] != e) ||
          (grapht[e] != -1 && grapht[e] != b)) {
        cout << "NO";
        exit(0);
      }
      graph[b] = e;
      grapht[e] = b;
    }
    for (auto i = s.begin(); i != s.end(); i++) {
      letter.insert(*i - 'a');
    }
  }
  for (int i = 0; i < 27; i++) {
    dfsn(i);
  }
  for (int i = 26; i >= 0; i--) {
    if (letter.find(i) != letter.end() && grapht[i] == -1) {
      dfs(i);
    }
  }
  reverse(result.begin(), result.end());
  cout << result;
  return 0;
}
