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;
using ll = long long;
using vi = vector<ll>;
using vvi = vector<vi>;
vvi g, gt;
string ans;
void dfs(int v) {
  if (count(ans.begin(), ans.end(), char('a' + v))) {
    cout << "NO";
    exit(0);
  }
  ans += char('a' + v);
  if (g[v].size()) dfs(g[v][0]);
}
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  set<pair<int, int>> edges;
  int n;
  cin >> n;
  string s;
  vi use(26, 0);
  for (int i = 0; i < n; i++) {
    cin >> s;
    for (int j = 1; j < s.size(); j++) {
      edges.insert({s[j - 1] - 'a', s[j] - 'a'});
    }
    vi cnt(26, 0);
    for (int j = 0; j < s.size(); j++) {
      if (++cnt[s[j] - 'a'] > 1) {
        return cout << "NO", 0;
      }
      use[s[j] - 'a'] = 1;
    }
  }
  g.resize(26);
  gt.resize(26);
  for (auto u : edges) {
    g[u.first].push_back(u.second);
    gt[u.second].push_back(u.first);
  }
  vi st;
  for (int i = 0; i < 26; i++) {
    if (g[i].size() > 1) return cout << "NO", 0;
    if (gt[i].size() > 1) return cout << "NO", 0;
    if (use[i] && !gt[i].size()) st.push_back(i);
  }
  sort(st.begin(), st.end());
  for (auto i : st) dfs(i);
  for (int i = 0; i < 26; i++)
    if (use[i] && count(ans.begin(), ans.end(), char('a' + i)) != 1)
      return cout << "NO", 0;
  cout << ans;
}
