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;
map<char, int> used, cnt;
map<char, char> next_ch;
string ans;
bool flag = false;
void dfs(char c) {
  if (used[c]) {
    flag = true;
    return;
  }
  ans.push_back(c);
  used[c] = 1;
  if (next_ch[c]) dfs(next_ch[c]);
  return;
}
int main() {
  ios::sync_with_stdio(0);
  int n;
  cin >> n;
  set<char> has;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int len = s.size();
    for (int i = 0; i < len; i++) {
      if (i != len - 1) {
        char c = s[i];
        if (next_ch[c] && next_ch[c] != s[i + 1]) {
          cout << "NO";
          return 0;
        }
        next_ch[c] = s[i + 1];
        cnt[s[i + 1]]++;
      }
      has.insert(s[0]);
    }
  }
  set<char>::iterator it = has.begin();
  for (; it != has.end(); it++) {
    if (!cnt[*it]) {
      dfs(*it);
      if (flag) {
        cout << "NO";
        return 0;
      }
    }
  }
  if (used.size() != cnt.size()) {
    cout << "NO";
    return 0;
  }
  cout << ans;
  return 0;
}
