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, through[30], nxt[30], prv[30], ever[30], ap[30];
char sir[100009];
void giveUp() {
  printf("NO\n");
  exit(0);
}
int main() {
  scanf("%d\n", &N);
  for (int i = 0; i < 26; i++) nxt[i] = prv[i] = -1;
  for (int i = 1; i <= N; i++) {
    int l;
    gets(sir + 1), l = strlen(sir + 1);
    for (int j = 1; j <= l; j++) sir[j] -= 'a';
    for (int j = 1; j < l; j++) {
      if (nxt[sir[j]] != sir[j + 1] && nxt[sir[j]] != -1)
        giveUp();
      else
        nxt[sir[j]] = sir[j + 1];
      if (ap[sir[j]]) giveUp();
      ap[sir[j]] = 1;
    }
    for (int j = l; j > 1; j--) {
      if (prv[sir[j]] != sir[j - 1] && prv[sir[j]] != -1)
        giveUp();
      else
        prv[sir[j]] = sir[j - 1];
    }
    if (ap[sir[l]]) giveUp();
    for (int j = 1; j <= l; j++) ap[sir[j]] = 0;
    for (int j = 1; j <= l; j++) ever[sir[j]] = 1;
  }
  vector<string> ans;
  for (int i = 0; i < 26; i++)
    if (ever[i] && prv[i] == -1) {
      string curr;
      int j = i;
      while (j != -1) curr.push_back('a' + j), through[j] = 1, j = nxt[j];
      ans.push_back(curr);
    }
  for (int i = 0; i < 26; i++)
    if (through[i] == 0 && ever[i] == 1) giveUp();
  sort(ans.begin(), ans.end());
  for (auto s : ans) printf("%s", s.c_str());
  printf("\n");
  return 0;
}
