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() {
  long long n;
  cin >> n;
  string s;
  vector<char> path(26, '.');
  vector<char> pathr(26, '.');
  vector<bool> have(26, 0);
  for (long long i = 0; i < n; i++) {
    string t;
    cin >> t;
    if (t.size() > 26) {
      cout << "NO";
      return 0;
    }
    vector<bool> curp(26, 0);
    long long pos = -1;
    for (auto &x : t) {
      if (curp[x - 'a']) {
        cout << "NO";
        return 0;
      }
      curp[x - 'a'] = 1;
      have[x - 'a'] = 1;
    }
    for (long long i = 1; i < t.size(); i++) {
      if (path[t[i - 1] - 'a'] == '.')
        path[t[i - 1] - 'a'] = t[i];
      else if (path[t[i - 1] - 'a'] != t[i]) {
        cout << "NO";
        return 0;
      }
      if (pathr[t[i] - 'a'] == '.')
        pathr[t[i] - 'a'] = t[i - 1];
      else if (pathr[t[i] - 'a'] != t[i - 1]) {
        cout << "NO";
        return 0;
      }
    }
  }
  bool flag = 0;
  vector<bool> used(26, 0);
  for (long long i = 0; i < 26; i++) {
    if (have[i] && !used[i] && pathr[i] == '.') {
      flag = 1;
      long long cur = i;
      while (path[cur] != '.') {
        s.push_back(cur + 'a');
        used[cur] = 1;
        cur = path[cur] - 'a';
        if (used[cur]) {
          cout << "NO";
          return 0;
        }
      }
      used[cur] = 1;
      s.push_back(cur + 'a');
    }
  }
  for (long long i = 0; i < 26; i++) {
    if (!used[i] && (pathr[i] != '.' || path[i] != '.')) {
      cout << "NO";
      return 0;
    }
  }
  if (!flag) {
    cout << "NO";
    return 0;
  }
  cout << s;
  cin >> n;
  return 0;
}
