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 N = 1e5 + 5;
int nxt[26];
int in[26];
int present[26];
int done[26];
char s[N];
int visited[26];
int get() {
  for (int i = 0; i < 26; i++) {
    if (present[i] && !visited[i] && in[i] == 0) {
      return i;
    }
  }
  return -1;
}
int valid() {
  for (int i = 0; i < 26; i++) {
    if (done[i] != present[i]) {
      return 0;
    }
  }
  return 1;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < 26; i++) {
    nxt[i] = -1;
  }
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    vector<int> tmp(26, 0);
    present[s[0] - 'a'] = 1;
    for (int j = 1; s[j]; j++) {
      int aa = s[j - 1] - 'a';
      int bb = s[j] - 'a';
      present[bb] = 1;
      tmp[bb]++;
      if (tmp[bb] > 1) {
        cout << "NO" << endl;
        return 0;
      }
      if (nxt[aa] >= 0 && nxt[aa] != bb) {
        cout << "NO" << endl;
        return 0;
      }
      in[bb] = 1;
      nxt[aa] = bb;
    }
  }
  int cc = -1;
  vector<string> ans;
  while ((cc = get()) != -1) {
    string s;
    s = 'a' + cc;
    visited[cc] = 1;
    cc = nxt[cc];
    while (cc != -1) {
      if (visited[cc]) {
        cout << "NO" << endl;
        return 0;
      }
      visited[cc] = 1;
      s += 'a' + cc;
      cc = nxt[cc];
    }
    ans.push_back(s);
  }
  for (auto e : ans) {
    for (auto c : e) {
      done[c - 'a'] = 1;
    }
  }
  if (!valid()) {
    cout << "NO" << endl;
    return 0;
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) {
    for (int j = 0; j < ans[i].length(); j++) {
      printf("%c", ans[i][j]);
    }
  }
  cout << endl;
}
