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 n, cnt[26];
vector<string> V;
string str[100010];
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> str[i];
  sort(str, str + n,
       [&](string a, string b) { return a.length() > b.length(); });
  auto chk = [&](string s) {
    static int c[26];
    memset(c, 0, sizeof(c));
    for (char ch : s) c[ch - 'a']++;
    for (int i = 0; i < 26; i++)
      if (c[i] > 1) return 0;
    return 1;
  };
  for (int _ = 0; _ < n; _++) {
    string s = str[_];
    if (s.length() > 26) cout << "NO\n", exit(0);
    if (!chk(s)) cout << "NO\n", exit(0);
    bool flag = 1;
    for (int i = 0; i < V.size(); i++) {
      if (~V[i].find(s)) {
        flag = 0;
        break;
      }
      memset(cnt, 0, sizeof(cnt));
      for (char c : s) cnt[c - 'a']++;
      for (char c : V[i]) cnt[c - 'a']++;
      bool foo = 0;
      for (int j = 0; j < 26; j++) {
        if (cnt[j] > 1) {
          foo = 1;
          break;
        }
      }
      if (!foo) continue;
      int pos = 0;
      while (pos < s.length() && cnt[s[pos] - 'a'] > 1) pos++;
      string t = V[i] + s.substr(pos);
      if (chk(t) && ~t.find(s)) {
        s = t, V.erase(V.begin() + i), i--;
      } else {
        pos = 0;
        while (pos < V[i].length() && cnt[V[i][pos] - 'a'] > 1) pos++;
        t = s + V[i].substr(pos);
        if (chk(t) && ~t.find(V[i]))
          s = t, V.erase(V.begin() + i), i--;
        else
          cout << "NO\n", exit(0);
      }
    }
    if (flag) V.push_back(s);
  }
  sort(V.begin(), V.end(), [&](string a, string b) { return a + b < b + a; });
  for (string s : V) cout << s;
  return 0;
}
