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 us[100], m[100], ous[100], om[100], pos[100], used[100];
bool cmp(string a, string b) { return a < b; }
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  bool k = 0;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < s.size(); j++) {
      int g1 = s[j] - 'a' + 1;
      used[g1] = 1;
    }
    for (int j = 0; j < s.size() - 1; j++) {
      int g1 = s[j] - 'a' + 1, g2 = s[j + 1] - 'a' + 1;
      if (us[g1] && m[g1] != g2) {
        k = 1;
      }
      if (ous[g2] && om[g2] != g1) {
        k = 1;
      }
      us[g1] = 1;
      m[g1] = g2;
      ous[g2] = 1;
      om[g2] = g1;
    }
  }
  if (k) {
    cout << "NO";
    return 0;
  }
  vector<string> v;
  for (int i = 0; i < 26; i++) {
    string ans = "";
    char t = i + 1;
    if (pos[t]) continue;
    while (ous[t]) {
      if (pos[t]) {
        cout << "NO";
        return 0;
      }
      pos[t] = 1;
      t = om[t];
    }
    while (us[t]) {
      pos[t] = 1;
      char t2 = t - 1 + 'a';
      ans += t2;
      t = m[t];
    }
    if (used[t]) {
      pos[t] = 1;
      char t2 = t - 1 + 'a';
      ans += t2;
    }
    if (!ans.empty()) {
      v.push_back(ans);
    }
  }
  sort(v.begin(), v.end(), cmp);
  for (int i = 0; i < v.size(); i++) {
    cout << v[i];
  }
}
