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;
char nex[256];
bool babu[256];
bool ada[256];
string str;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> str;
    for (int j = 0; j < (int)str.length() - 1; j++) {
      ada[(int)str[j]] = 1;
      if (nex[(int)str[j]] != 0 && nex[(int)str[j]] != str[j + 1]) {
        printf("NO\n");
        return 0;
      } else {
        nex[(int)str[j]] = str[j + 1];
        babu[(int)str[j + 1]] = true;
      }
    }
    ada[(int)str[(int)str.length() - 1]] = true;
  }
  int itung = 0;
  for (char x = 'a'; x <= 'z'; x++) {
    if (ada[(int)x]) itung++;
  }
  string jaw = "";
  for (char i = 'a'; i <= 'z'; i++) {
    char x = i;
    if (!babu[(int)x] && ada[(int)x]) {
      while ((int)x != 0) {
        jaw += x;
        x = nex[(int)x];
        if ((int)x != 0 && !babu[(int)x]) {
          printf("NO\n");
          return 0;
        }
        babu[(int)x] = false;
        ada[(int)x] = false;
      }
    }
  }
  if (itung != (int)jaw.length()) {
    jaw = "NO";
  }
  cout << jaw << "\n";
}
