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>
const long long INF = 4e18L + 1;
const int IINF = 2e9 + 1;
using namespace std;
template <class TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char* sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
bool wspolny(string s1, string s2) {
  for (auto e1 : s1)
    for (auto e2 : s2)
      if (e1 == e2) return true;
  return false;
}
bool ok = true;
string merge(string s1, string s2) {
  char c = 'z' + 1;
  string res;
  for (int i = 0; i < (int)((s1).size()) + (int)((s2).size()) + 1; i++)
    res += c;
  int ind = -1;
  for (int i = 0; i < (int)((s1).size()); i++) {
    if (s1[i] == s2[0]) ind = i;
  }
  if (ind == -1) {
    swap(s1, s2);
    for (int i = 0; i < (int)((s1).size()); i++) {
      if (s1[i] == s2[0]) ind = i;
    }
  }
  if (ind == -1) {
    ok = false;
    return res;
  }
  for (int i = 0; i < (int)((s1).size()); i++) res[i] = s1[i];
  for (int i = 0; i < (int)((s2).size()); i++) res[i + ind] = s2[i];
  if (res.find(s1) == string::npos) ok = false;
  while (res.back() == c) res.pop_back();
  return res;
}
void sprawdz(string s) {
  if ((int)((s).size()) > 30) {
    cout << "NO";
    exit(0);
  }
  sort((s).begin(), (s).end());
  for (int i = 1; i < (int)((s).size()); i++)
    if (s[i] == s[i - 1]) {
      cout << "NO\n";
      exit(0);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  vector<string> V(1);
  cin >> V[0];
  sprawdz(V[0]);
  for (int i = 1; i < n; i++) {
    string nowy;
    cin >> nowy;
    sprawdz(nowy);
    vector<string> NV(1, nowy);
    for (auto& el : V) {
      if (wspolny(NV[0], el))
        NV[0] = merge(NV[0], el);
      else
        NV.push_back(el);
    }
    V = NV;
    if (ok == false) {
      cout << "NO";
      return 0;
    }
  }
  for (auto& el : V) sprawdz(el);
  if (ok == false) {
    cout << "NO";
    return 0;
  }
  sort((V).begin(), (V).end());
  for (auto& el : V) cout << el;
  return 0;
}
