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 N = 1e5 + 20;
const int MOD = 1e9 + 7;
const int alp = 26;
int v[alp][alp], h[alp];
int main() {
  int n;
  cin >> n;
  memset(h, -1, sizeof(h));
  while (n--) {
    string s;
    cin >> s;
    for (int i = 0; i < s.length() - 1; i++) v[s[i] - 'a'][s[i + 1] - 'a'] = 1;
    for (auto ch : s) h[ch - 'a'] = 0;
  }
  string ans;
  for (int st = 0; st < alp; st++)
    if (h[st] != -1) {
      int n1 = 0;
      for (int i = 0; i < alp; i++) n1 += v[i][st];
      if (n1) continue;
      int cur = st;
      while (1) {
        h[cur] = 1;
        ans.push_back(cur + 'a');
        int deg = count(v[cur], v[cur] + alp, 1);
        if (deg > 1) {
          cout << "NO";
          return 0;
        }
        if (!deg) break;
        int ncur = 0;
        while (!v[cur][ncur]) ncur++;
        if (h[ncur]) {
          cout << "NO";
          return 0;
        }
        cur = ncur;
      }
    }
  if (count(h, h + alp, 0)) {
    cout << "NO";
    return 0;
  }
  cout << ans;
}
