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;
template <class X, class Y>
void minimize(X &x, const Y &y) {
  if (x > y) x = y;
}
template <class X, class Y>
void maximize(X &x, const Y &y) {
  if (x < y) x = y;
}
template <class T>
T Abs(const T &x) {
  return (x < 0 ? -x : x);
}
const string NO = "NO";
void answerNO(void) {
  cout << NO << endl;
  exit(EXIT_SUCCESS);
}
bool havePath[226][226], need[226][226];
int countIn[226], countOut[226];
bool compare(const string &s, const string &t) { return s + t < t + s; }
void process(void) {
  int needChar = 0;
  int n;
  cin >> n;
  for (int love = 0, _n = (n); love < _n; love = love + 1) {
    string s;
    cin >> s;
    int mask = 0;
    for (__typeof((s).begin()) it = (s).begin(); it != (s).end(); it++) {
      if ((((mask) >> (*it - 'a')) & 1)) answerNO();
      mask |= (1LL << (*it - 'a'));
    }
    for (int i = 0, _n = ((int)s.size() - 1); i < _n; i = i + 1) {
      int x = s[i] - 'a';
      int y = s[i + 1] - 'a';
      need[x][y] = true;
    }
    needChar |= mask;
  }
  for (int i = 0, _n = (26); i < _n; i = i + 1) havePath[i][i] = true;
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    for (int j = 0, _n = (26); j < _n; j = j + 1) havePath[i][j] |= need[i][j];
  for (int k = 0, _n = (26); k < _n; k = k + 1)
    for (int i = 0, _n = (26); i < _n; i = i + 1)
      for (int j = 0, _n = (26); j < _n; j = j + 1)
        havePath[i][j] |= havePath[i][k] && havePath[k][j];
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    for (int j = 0, _n = (26); j < _n; j = j + 1)
      if (i != j && havePath[i][j] && havePath[j][i]) answerNO();
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    for (int j = 0, _n = (26); j < _n; j = j + 1)
      if (need[i][j]) {
        countIn[j]++;
        countOut[i]++;
      }
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    if (countIn[i] > 1 || countOut[i] > 1) answerNO();
  vector<string> str;
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    if (countOut[i] > 0 && countIn[i] == 0) {
      int tmp = i;
      string s;
      while (countOut[tmp] > 0) {
        s.push_back(tmp + 'a');
        for (int j = 0, _n = (26); j < _n; j = j + 1)
          if (need[tmp][j]) {
            tmp = j;
            break;
          }
      }
      s.push_back(tmp + 'a');
      str.push_back(s);
    }
  for (int i = 0, _n = (26); i < _n; i = i + 1)
    if ((((needChar) >> (i)) & 1) && countIn[i] == 0 && countOut[i] == 0)
      str.push_back(string(1, i + 'a'));
  sort((str).begin(), (str).end(), compare);
  for (__typeof((str).begin()) it = (str).begin(); it != (str).end(); it++)
    cout << *it;
  cout << endl;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  process();
  return 0;
}
