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 main() {
  int n;
  scanf("%d", &n);
  int next[50] = {0}, in[50] = {0}, book[50] = {0}, front[50] = {0}, cnt = 0,
      ans[50], total = 0;
  memset(next, -1, sizeof(next));
  memset(front, -1, sizeof(front));
  char s[200005];
  while (n--) {
    scanf("%s", s);
    int len = strlen(s);
    if (len == 1) {
      book[s[0] - 'a' + 1] = 1;
    }
    for (int i = 0; i <= len - 2; i++) {
      int a = s[i] - 'a' + 1;
      int b = s[i + 1] - 'a' + 1;
      book[a] = 1;
      book[b] = 1;
      if (next[a] != b && next[a] != -1) {
        printf("NO\n");
        return 0;
      }
      if (front[b] != a && front[b] != -1) {
        printf("NO\n");
        return 0;
      }
      next[a] = b;
      front[b] = a;
      in[b]++;
    }
  }
  for (int i = 1; i <= 26; i++) {
    if (book[i]) {
      total++;
      if (in[i] == 0) {
        for (int j = i; j != -1; j = next[j]) {
          ans[++cnt] = j;
        }
      }
    }
  }
  if (cnt == total) {
    for (int i = 1; i <= cnt; i++) printf("%c", (char)(ans[i] + 'a' - 1));
    printf("\n");
  } else {
    printf("NO\n");
    return 0;
  }
  return 0;
}
