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 = 100005;
const int A = 26;
int n;
string s[N];
int cnt[A];
bool cmp(const string& x, const string& y) {
  return ((int)x.size()) > ((int)y.size());
}
void die() {
  cout << "NO\n";
  exit(0);
}
vector<int> g[N];
int was[A], have[A], in[A];
string result;
void dfs(int v) {
  was[v] = 1;
  result += char(v + 'a');
  for (auto& to : g[v]) {
    if (!was[to]) dfs(to);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> s[i];
    for (auto& c : s[i]) ++cnt[c - 'a'], have[c - 'a'] = 1;
    for (int j = 0; j < (A); ++j) {
      if (cnt[j] > 1) die();
      cnt[j] = 0;
    }
    for (int j = 1; j < ((int)s[i].size()); ++j) {
      g[s[i][j - 1] - 'a'].push_back(s[i][j] - 'a');
      ++in[s[i][j] - 'a'];
    }
  }
  for (int i = 0; i < (A); ++i)
    if (!was[i] && !in[i] && have[i]) dfs(i);
  for (int i = 0; i < (n); ++i)
    if (result.find(s[i]) == string::npos) die();
  cout << result << '\n';
  return 0;
}
