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;
string s;
int nx[26], bk[26], app[26], used[26], vis[26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  fill(nx, nx + 26, -1);
  fill(bk, bk + 26, -1);
  cin >> n;
  while (n--) {
    cin >> s;
    for (int i = 0; i < s.size() - 1; i++) {
      int x = s[i] - 'a', y = s[i + 1] - 'a';
      vis[x] = true;
      if ((nx[x] != y && nx[x] != -1) || (x != bk[y] && bk[y] != -1)) {
        cout << "NO";
        return 0;
      } else {
        nx[s[i] - 'a'] = s[i + 1] - 'a';
        bk[s[i + 1] - 'a'] = s[i] - 'a';
      }
    }
    vis[s.back() - 'a'] = true;
    if (s.size() == 1 && !app[s[0] - 'a'])
      app[s[0] - 'a'] = 1;
    else if (s.size() == 1) {
      cout << "NO";
      return 0;
    }
  }
  vector<string> res;
  bool flag = false;
  for (int i = 0; i < 26; i++) {
    string x;
    if (bk[i] == -1 && nx[i] != -1) {
      flag = true;
      int c = i;
      do {
        x.push_back(c + 'a');
        used[c] = 1;
        c = nx[c];
      } while (c != -1);
    } else if (app[i] && bk[i] == -1 && nx[i] == -1) {
      used[i] = true;
      flag = true;
      x.push_back(i + 'a');
    }
    res.push_back(x);
  }
  for (int i = 0; i < 26; i++)
    if (!used[i] && vis[i]) {
      cout << "NO";
      return 0;
    }
  if (!flag) {
    cout << "NO";
    return 0;
  }
  sort(res.begin(), res.end());
  for (string x : res) cout << x;
}
