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;
vector<bool> used;
vector<vector<int> > g, rg;
vector<int> ord;
int main() {
  int n;
  cin >> n;
  vector<string> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  g.resize(26, vector<int>(26)), rg.resize(26, vector<int>(26));
  used.resize(26);
  vector<bool> uc(26);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)a[i].size(); j++) {
      uc[a[i][j] - 'a'] = 1;
      if (j == (int)a[i].size() - 1) break;
      char c1 = a[i][j];
      char c2 = a[i][j + 1];
      g[c1 - 'a'][c2 - 'a'] = 1;
      rg[c2 - 'a'][c1 - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    int cnt = count(g[i].begin(), g[i].end(), 1);
    if (cnt > 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  string ans;
  int cnt = 0;
  for (int i = 0; i < 26; i++) {
    if (!uc[i]) continue;
    cnt++;
    if (count(rg[i].begin(), rg[i].end(), 1) > 0) continue;
    int u = i;
    while (1) {
      ans += (u + 'a');
      used[u] = 1;
      int pos = find(g[u].begin(), g[u].end(), 1) - g[u].begin();
      if (pos == 26) break;
      if (used[pos]) {
        cout << "NO" << endl;
        return 0;
      }
      u = pos;
    }
  }
  if (ans.size() != cnt) {
    cout << "NO" << endl;
    return 0;
  }
  cout << ans << endl;
  return 0;
}
