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() {
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < n; ++i) cin >> s[i];
  vector<int> use(26);
  for (int i = 0; i < n; ++i) {
    int cnt[26] = {};
    for (char c : s[i]) {
      cnt[c - 'a']++;
      use[c - 'a'] = 1;
    }
    for (int j = 0; j < 26; ++j) {
      if (cnt[j] >= 2) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  vector<int> next(26, -1), past(26, -1);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < s[i].length() - 1; ++j) {
      next[s[i][j] - 'a'] = s[i][j + 1] - 'a';
      past[s[i][j + 1] - 'a'] = s[i][j] - 'a';
    }
  }
  vector<string> cand;
  for (int i = 0; i < 26; ++i) {
    if (!use[i] || past[i] != -1) continue;
    string tmp;
    int x = i;
    while (x != -1) {
      if (next[x] != -1 && past[next[x]] != x) {
        cout << "NO" << endl;
        return 0;
      }
      tmp += ('a' + x);
      x = next[x];
    }
    cand.push_back(tmp);
  }
  sort((cand).begin(), (cand).end());
  string ans;
  for (string str : cand) ans += str;
  for (int i = 0; i < n; ++i) {
    if (ans.find(s[i]) == string::npos) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << ans << endl;
  return 0;
}
