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;
const int maxn = 1e6 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
set<int> G[38], RG[38];
int have[38];
int n;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  string t;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    t += s;
    int N = s.length();
    for (char c : s) have[c - 'a'] = 1;
    for (int j = 0; j < N - 1; j++) {
      G[s[j] - 'a'].insert(s[j + 1] - 'a');
      if (((int)(G[s[j] - 'a']).size()) >= 2) return cout << "NO" << endl, 0;
      RG[s[j + 1] - 'a'].insert(s[j] - 'a');
      if (((int)(RG[s[j + 1] - 'a']).size()) >= 2)
        return cout << "NO" << endl, 0;
    }
  }
  priority_queue<int, vector<int>, greater<int> > pq;
  for (int i = 0; i < 26; i++) {
    if (((int)(RG[i]).size()) == 0 && have[i]) pq.push(i);
  }
  string ans;
  while (!pq.empty()) {
    int now = pq.top();
    pq.pop();
    while (1) {
      ans += (char)(now + 'a');
      if (((int)(G[now]).size()) == 0) break;
      now = (*(G[now].begin()));
      if (ans.length() > 1000) return cout << "NO" << endl, 0;
    }
  }
  set<int> sss;
  for (char c : t) sss.insert(c);
  if (ans.length() != sss.size())
    cout << "NO" << endl;
  else
    cout << ans << endl;
}
