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;
int n;
string s[100005];
int a[50][50];
int f[50], son[50];
int vis[50];
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> s[i];
  memset(a, 0, sizeof(a));
  memset(f, -1, sizeof(f));
  memset(son, -1, sizeof(son));
  memset(vis, 0, sizeof(vis));
  for (int i = 0; i < n; i++) {
    int len = s[i].size();
    vis[s[i][0] - 'a'] = 1;
    for (int j = 1; j < len; j++) {
      int q = s[i][j] - 'a', w = s[i][j - 1] - 'a';
      f[q] = w;
      son[w] = q;
      a[w][q] = 1;
      vis[q] = vis[w] = 1;
    }
  }
  string s1;
  int ok[50];
  memset(ok, 0, sizeof(ok));
  int flag = 1;
  for (int i = 0; i < 26; i++) {
    if (ok[i] == 0 && f[i] == -1 && vis[i] == 1) {
      int j = i;
      while (j != -1 && ok[j] == 0) {
        s1 += (j + 'a');
        ok[j] = 1;
        j = son[j];
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (s1.find(s[i]) == s1.npos) flag = 0;
  if (flag)
    cout << s1 << endl;
  else
    cout << "NO" << endl;
  return 0;
}
