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;
void die() {
  cout << "NO";
  exit(0);
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  bool ins[26] = {};
  bool vis[26] = {};
  int next[26];
  int prev[26];
  memset(next, -1, sizeof next);
  memset(prev, -1, sizeof prev);
  string sub;
  while (n--) {
    cin >> sub;
    int m = sub.size();
    if (m > 26) die();
    for (char c : sub) ins[c - 'a'] = true, vis[c - 'a'] = false;
    for (int i = 0; i < m; ++i) {
      int idx = sub[i] - 'a';
      if (vis[idx]) die();
      vis[idx] = true;
      if (0 < i) {
        int prev_idx = sub[i - 1] - 'a';
        if (prev[idx] != -1 && prev_idx != prev[idx]) die();
        prev[idx] = prev_idx;
      }
      if (i + 1 < m) {
        int next_idx = sub[i + 1] - 'a';
        if (next[idx] != -1 && next_idx != next[idx]) die();
        next[idx] = next_idx;
      }
    }
  }
  string ans;
  ans.reserve(32);
  for (int i = 0; i < 26; ++i) {
    if (not ins[i] || prev[i] != -1) continue;
    for (int j = i; j >= 0; j = next[j]) ans.push_back(char(j + 'a'));
  }
  if (ans.empty()) die();
  for (char c : ans) ins[c - 'a'] = vis[c - 'a'] = false;
  for (char c : ans) {
    int idx = c - 'a';
    if (vis[idx]) die();
    vis[idx] = true;
  }
  if (count(ins, ins + 26, true)) die();
  cout << ans;
}
