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 INF = 0x3f3f3f3f;
const long long MOD = 1e9 + 7;
const double PI = acos(-1.0);
string str[101000];
int take[30];
int Next[30];
int vis[30];
string res[400];
int main() {
  int n;
  cin >> n;
  if (n > 27 * 13) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> str[i];
    memset(take, 0, sizeof take);
    for (int j = 0; j < str[i].length(); ++j) {
      if (!take[str[i][j] - 'a'])
        take[str[i][j] - 'a'] = 1;
      else {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  memset(vis, 0, sizeof vis);
  memset(take, 0, sizeof take);
  memset(Next, -1, sizeof(Next));
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < str[i].length() - 1; ++j) {
      int x = str[i][j] - 'a';
      int y = str[i][j + 1] - 'a';
      if (take[x]) {
        if (Next[x] != y) {
          if (Next[x] != -1) {
            cout << "NO" << endl;
            return 0;
          } else
            Next[x] = y;
        }
      } else {
        take[x] = 1;
        Next[x] = y;
      }
    }
    for (int j = 1; j < str[i].length(); ++j) vis[str[i][j] - 'a']++;
    take[str[i][str[i].length() - 1] - 'a'] = 1;
  }
  int p = 0;
  for (int i = 0; i <= 25; ++i) {
    if (vis[i] == 0 && take[i] == 1) {
      res[++p].push_back(i + 'a');
      int now = Next[i];
      int cnt = 0;
      while (now != -1 && cnt < 50)
        res[p].push_back(now + 'a'), cnt++, now = Next[now];
      if (cnt >= 30) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  int cc = 0;
  for (int i = 0; i <= 25; ++i)
    if (take[i]) cc++;
  sort(res + 1, res + p + 1);
  string ans;
  ans.clear();
  for (int i = 1; i <= p; ++i) ans += res[i];
  memset(take, 0, sizeof take);
  for (int i = 0; i < ans.length(); ++i) {
    if (!take[ans[i] - 'a'])
      take[ans[i] - 'a'] = 1;
    else {
      cout << "NO" << endl;
      return 0;
    }
  }
  if (cc != ans.length()) {
    cout << "NO" << endl;
    return 0;
  }
  cout << ans << endl;
}
