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;
const int N = 100000;
const int inf = (int)1e9 + 1;
const long long big = (long long)1e18 + 1;
const int P = 239;
const int MOD = (int)1e9 + 7;
const int MOD1 = (int)1e9 + 9;
const double eps = 1e-9;
const double pi = atan2(0, -1);
const int ABC = 26;
string a[N];
int cnt[N][26];
bool used[26];
set<int> g[26], gr[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(20);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < (int)a[i].size(); j++) {
      cnt[i][a[i][j] - 'a']++;
      used[a[i][j] - 'a'] = true;
      if (cnt[i][a[i][j] - 'a'] == 2) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < (int)a[i].size() - 1; j++) {
      g[a[i][j] - 'a'].insert(a[i][j + 1] - 'a');
      gr[a[i][j + 1] - 'a'].insert(a[i][j] - 'a');
    }
  for (int i = 0; i < 26; i++)
    if ((int)g[i].size() > 1) {
      cout << "NO\n";
      return 0;
    }
  vector<string> res;
  for (int i = 0; i < 26; i++)
    if (used[i] && (int)gr[i].size() == 0) {
      string t;
      int cur = i;
      while (true) {
        used[cur] = false;
        t += (char)('a' + cur);
        if ((int)g[cur].size() == 0)
          break;
        else {
          if (used[*g[cur].begin()])
            cur = *g[cur].begin();
          else {
            cout << "NO\n";
            return 0;
          }
        }
      }
      res.push_back(t);
    }
  for (int i = 0; i < 26; i++)
    if (used[i]) {
      cout << "NO\n";
      return 0;
    }
  sort(res.begin(), res.end());
  for (auto it : res) cout << it;
  cout << "\n";
  return 0;
}
