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;
int n;
string s[N];
int cnt[N][26];
bool used[26];
set<int> g[26], b[26];
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> s[i];
    for (int j = 0; j < s[i].size(); ++j) {
      cnt[i][s[i][j] - 'a']++;
      used[s[i][j] - 'a'] = true;
      if (cnt[i][s[i][j] - 'a'] > 1) {
        cout << "NO";
        return 0;
      }
    }
  }
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < s[i].size() - 1; ++j) {
      g[s[i][j] - 'a'].insert(s[i][j + 1] - 'a');
      b[s[i][j + 1] - 'a'].insert(s[i][j] - 'a');
    }
  for (int i = 0; i < 26; ++i) {
    if (g[i].size() > 1) {
      cout << "NO";
      return 0;
    }
  }
  vector<string> res;
  for (int i = 0; i < 26; ++i) {
    if (used[i] && b[i].size() == 0) {
      string t;
      int cur = i;
      while (1) {
        used[cur] = false;
        t += (char)('a' + cur);
        if (g[cur].size() == 0)
          break;
        else {
          if (used[*g[cur].begin()])
            cur = *g[cur].begin();
          else {
            cout << "NO";
            return 0;
          }
        }
      }
      res.push_back(t);
    }
  }
  for (int i = 0; i < 26; ++i)
    if (used[i]) {
      cout << "NO";
      return 0;
    }
  sort(res.begin(), res.end());
  for (auto v : res) cout << v;
  return 0;
}
