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;
string str, res;
int n, from[26], to[26], vis[26], used[26];
bool chk(int x, int y) {
  if (to[x] == -1) to[x] = y;
  if (from[y] == -1) from[y] = x;
  return (to[x] == y && from[y] == x);
}
void solve(int pos) {
  if (pos == -1) return;
  vis[pos] = 1;
  res += pos + 97;
  solve(to[pos]);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  fill(from, from + 26, -1);
  fill(to, to + 26, -1);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> str;
    for (auto x : str) used[x - 97] = 1;
    for (int j = 0; j + 1 < str.length(); j++) {
      int x = str[j] - 97, y = str[j + 1] - 97;
      if (chk(x, y))
        ;
      else {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (used[i] && from[i] == -1) solve(i);
  }
  for (int i = 0; i < 26; i++)
    if (used[i] && not vis[i]) {
      cout << "NO" << endl;
      return 0;
    }
  cout << res << endl;
  return 0;
}
