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, ans;
bool b[1000], pred[1000];
int a[1000];
int main() {
  ios_base ::sync_with_stdio(false);
  int i, j;
  for (i = 0; i <= 800; i++) {
    b[i] = false;
    pred[i] = false;
    a[i] = -1;
  }
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> s;
    for (j = 0; j < s.length() - 1; j++) {
      int v = (int)s[j];
      int to = (int)s[j + 1];
      b[v] = b[to] = true;
      pred[to] = true;
      if (a[v] > -1 && a[v] != to) {
        cout << "NO";
        return 0;
      }
      a[v] = to;
    }
    b[(int)s[s.length() - 1]] = true;
  }
  for (i = 0; i <= 800; i++) {
    if (b[i] == true && pred[i] == false) {
      ans += (char)i;
      b[i] = false;
      int v = i;
      while (a[v] > -1) {
        v = a[v];
        if (b[v] == false) {
          cout << "NO";
          return 0;
        }
        ans += (char)v;
        b[v] = false;
      }
    }
  }
  for (i = 0; i <= 800; i++)
    if (b[i] == true) {
      cout << "NO";
      return 0;
    }
  cout << ans;
  return 0;
}
