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 anspos = 1;
string ans;
string alphabets = "abcdefghijklmnopqrstuvwxyz";
void dfs(int v, int visited[], set<int> graph[], set<int> &curstack) {
  visited[v] = 1;
  ans.push_back(alphabets[v]);
  curstack.insert(v);
  for (auto x : graph[v]) {
    if (curstack.find(x) != curstack.end()) anspos = 0;
    if (visited[x] == 0) {
      dfs(x, visited, graph, curstack);
    }
  }
  curstack.erase(v);
}
int32_t main() {
  int n;
  cin >> n;
  int flag = 0;
  vector<string> arr;
  int freq[26] = {0};
  set<int> graph[26];
  int visited[26], indegree[26] = {0};
  ;
  for (int i = 0; i < 26; i++) visited[i] = 1;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < 26; j++) freq[j] = 0;
    for (int j = 0; j < s.size(); j++) {
      freq[s[j] - 'a']++;
      visited[s[j] - 'a'] = 0;
    }
    for (int j = 0; j < 26; j++)
      if (freq[j] > 1) flag = 1;
    for (int i = 0; i < s.size() - 1; i++) {
      if (graph[s[i] - 'a'].find(s[i + 1] - 'a') == graph[s[i] - 'a'].end())
        indegree[s[i + 1] - 'a']++;
      graph[s[i] - 'a'].insert(s[i + 1] - 'a');
      if (graph[s[i] - 'a'].size() > 1 || indegree[s[i + 1] - 'a'] > 1)
        flag = 1;
    }
  }
  if (flag == 1) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 0; i < 26; i++) {
    if (visited[i] == 1 || indegree[i] != 0) continue;
    set<int> curstack;
    dfs(i, visited, graph, curstack);
  }
  for (int i = 0; i < 26; i++)
    if (visited[i] == 0) anspos = 0;
  if (anspos == 0)
    cout << "NO\n";
  else
    cout << ans << "\n";
}
