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;
int pro[27];
int cnt[100001][28];
string str[100001];
int cnt2[28];
bool gb[100001];
int cnt3[28];
bool comp(string a, string b) { return a + b < b + a; }
string append(string a, string b, int c) {
  int f = 0;
  for (int i = 0; i < a.size(); i++) {
    if ((a[i] - 'a') == c) f = i;
  }
  int t = 0;
  for (int i = 0; i < b.size(); i++) {
    if ((b[i] - 'a') == c) t = i;
  }
  for (int i = -min(f, t); i < min((int)a.size() - f, (int)b.size() - t); i++) {
    if (a[f + i] != b[t + i]) return "NO";
  }
  string ans = "";
  memset(cnt3, 0, sizeof(cnt3));
  for (int i = -max(f, t); i < max((int)a.size() - f, (int)b.size() - t); i++) {
    if (f + i >= 0 && f + i < a.size()) {
      ans += a[f + i];
    } else {
      ans += b[t + i];
    }
    cnt3[ans[(int)ans.size() - 1] - 'a']++;
    if (cnt3[ans[(int)ans.size() - 1] - 'a'] >= 2) return "NO";
  }
  return ans;
}
string solve() {
  memset(cnt2, -1, sizeof(cnt2));
  for (int i = 0; i < n; i++) {
    string tmp = str[i];
    for (int j = 0; j < tmp.size(); j++) {
      if (cnt2[tmp[j] - 'a'] != -1 && cnt2[tmp[j] - 'a'] != i) {
        int v = cnt2[tmp[j] - 'a'];
        str[i] = append(str[i], str[v], (tmp[j] - 'a'));
        str[v] = "";
        gb[v] = false;
        if (str[i] == "NO") return "NO";
        for (int k = 0; k < 26; k++) {
          if (cnt2[k] == v) cnt2[k] = i;
        }
      }
      cnt2[tmp[j] - 'a'] = i;
    }
    gb[i] = true;
  }
  vector<string> vec;
  for (int i = 0; i < n; i++) {
    if (gb[i]) {
      vec.push_back(str[i]);
    }
  }
  sort(vec.begin(), vec.end(), comp);
  string ans = "";
  for (int i = 0; i < vec.size(); i++) {
    ans += vec[i];
  }
  return ans;
}
int main(void) {
  scanf("%d", &n);
  bool flag = true;
  for (int i = 0; i < n; i++) {
    cin >> str[i];
    for (int j = 0; j < str[i].size(); j++) {
      cnt[i][(str[i][j] - 'a')]++;
      if (cnt[i][(str[i][j] - 'a')] >= 2) flag = false;
    }
  }
  if (!flag) {
    printf("NO\n");
    return 0;
  }
  cout << solve() << endl;
  return 0;
}
