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 maxn = 1e5;
int nxt[26] = {};
bool used[26] = {}, vis[26] = {}, inmid[26] = {};
bool run() {
  fill_n(nxt, 26, -1);
  int N;
  cin >> N;
  for (auto i = (0); i != (N); ++i) {
    string str;
    cin >> str;
    vector<bool> seen(26, false);
    for (char ch : str) {
      used[ch - 'a'] = true;
      if (seen[ch - 'a']) return false;
      seen[ch - 'a'] = true;
    }
    int len = str.size();
    for (int j = 1; j < len; j++) {
      int x0 = str[j - 1] - 'a';
      int x1 = str[j] - 'a';
      if (nxt[x0] < 0) nxt[x0] = x1, inmid[x1] = true;
      if (nxt[x0] != x1) return false;
    }
  }
  vector<int> ans;
  for (int i = 0; i < 26; i++) {
    if (!used[i] || vis[i] || inmid[i]) continue;
    for (int j = i; j >= 0; j = nxt[j]) {
      if (vis[j]) return false;
      vis[j] = true;
      ans.push_back(j);
    }
  }
  if (ans.empty()) return false;
  for (int i = 0; i < 26; i++) {
    if (used[i] && !vis[i]) return false;
  }
  for (int i : ans) cout << ((char)(i + 'a'));
  cout << endl;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  if (!run()) {
    cout << "NO" << endl;
  }
  return 0;
}
