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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> next(30, -1);
  vector<bool> check(26), flag(26);
  vector<int> pos;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    flag[(int)(s[0] - 'a')] = 1;
    for (int j = 0; j < s.length() - 1; j++) {
      int cur1 = (int)(s[j] - 'a'), cur2 = (int)(s[j + 1] - 'a');
      flag[cur1] = 1;
      flag[cur2] = 1;
      check[cur2] = 1;
      if (next[cur1] == -1) {
        next[cur1] = cur2;
      } else if (next[cur1] != cur2) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    vector<int> used2(26);
    int cur = i;
    used2[cur] = 1;
    while (next[cur] != -1) {
      cur = next[cur];
      if (used2[cur] == 1) {
        cout << "NO" << endl;
        return 0;
      }
      used2[cur] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (!check[i] && flag[i]) {
      pos.push_back(i);
    }
  }
  set<string> s;
  vector<bool> used(26);
  string tmp = "";
  for (int i = 0; i < pos.size(); i++) {
    if (used[pos[i]]) {
      continue;
    }
    int cur = pos[i];
    used[cur] = 1;
    tmp += (char)(cur + 'a');
    while (next[cur] != -1) {
      cur = next[cur];
      if (used[cur]) {
        cout << "NO" << endl;
        return 0;
      }
      tmp += (char)(cur + 'a');
      used[cur] = 1;
    }
    s.insert(tmp);
    tmp = "";
  }
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it;
  }
  cout << endl;
}
