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 s[100050];
int len;
char nxt[30], prv[30];
bool use[30];
bool vis[30];
int main() {
  int n;
  scanf("%d", &n);
  while (n--) {
    scanf("%s", s);
    len = strlen(s);
    for (int i = 0; i < len; i++) {
      int k = s[i] - 'a';
      if (i) {
        if (prv[k] && prv[k] != s[i - 1]) return puts("NO"), 0;
        prv[k] = s[i - 1];
      }
      if (i < len - 1) {
        if (nxt[k] && nxt[k] != s[i + 1]) return puts("NO"), 0;
        nxt[k] = s[i + 1];
      }
      use[k] = true;
    }
  }
  len = 0;
  for (int i = 0; i < 26; i++)
    if (use[i] && !prv[i]) {
      vis[i] = true;
      s[len++] = i + 'a';
      int j = nxt[i];
      while (true) {
        if (!j) break;
        s[len++] = j;
        if (vis[j - 'a']) return puts("NO"), 0;
        vis[j - 'a'] = true;
        j = nxt[j - 'a'];
      }
    }
  for (int i = 0; i < 26; i++)
    if (use[i] && !vis[i]) return puts("NO"), 0;
  puts(len ? s : "NO");
  return 0;
}
