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;
vector<string> a;
set<int> e[26];
set<int> u[26];
bool vis[26];
bool ima[26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cerr.tie(nullptr);
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    a.push_back(s);
  }
  bool ok = true;
  for (string s : a) {
    for (size_t i = 0; i + 1 < s.size(); i++) {
      e[s[i] - 'a'].insert(s[i + 1] - 'a');
      u[s[i + 1] - 'a'].insert(s[i] - 'a');
    }
    vector<int> br(26, 0);
    for (char x : s) {
      if (2 == ++br[x - 'a']) {
        ok = false;
      }
      ima[x - 'a'] = true;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (e[i].size() > 1 || u[i].size() > 1) {
      cout << "NO\n";
      return 0;
    }
  }
  if (!ok) {
    cout << "NO\n";
    return 0;
  }
  string sol;
  for (int i = 0; i < 26; i++) {
    if (!vis[i] && ima[i] && u[i].size() == 0) {
      int t = i;
      while (1) {
        vis[t] = true;
        sol += t + 'a';
        if (e[t].size() == 0) {
          break;
        } else {
          t = *e[t].begin();
        }
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (ima[i] && !vis[i]) {
      ok = false;
    }
  }
  if (!ok) {
    cout << "NO\n";
    return 0;
  }
  cout << sol << '\n';
}
