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;
template <class T>
inline bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
inline bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
int edge[26][26], h[26];
string ans;
int N;
void bye() {
  cout << "NO" << '\n';
  exit(0);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < (int)s.length() - 1; ++j) {
      edge[(int)s[j] - 'a'][(int)s[j + 1] - 'a'] = 1;
    }
    for (char ch : s) h[(int)ch - 'a'] = 1;
  }
  for (int alp = 0; alp < 26; ++alp) {
    if (!h[alp]) continue;
    int prev = 0;
    for (int i = 0; i < 26; ++i) prev += edge[i][alp];
    if (prev) continue;
    int cur_alp = alp;
    while (true) {
      h[cur_alp] = 2;
      ans += (char)('a' + cur_alp);
      int deg = count(edge[cur_alp], edge[cur_alp] + 26, 1);
      if (deg > 1) bye();
      if (deg == 0) break;
      int nxt = -1;
      for (int i = 0; i < 26; ++i)
        if (edge[cur_alp][i] == 1) nxt = i;
      if (h[nxt] == 2) bye();
      cur_alp = nxt;
    }
  }
  if (count(h, h + 26, 1)) bye();
  cout << ans << '\n';
}
