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;
const int maxn = 100010;
int n, len, vis[30], to[30], from[30], vis2[30], ans[maxn], cnt;
bool flag = true;
char s[maxn];
int main() {
  for (int i = 0; i < 26; i++) from[i] = to[i] = -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    len = strlen(s + 1);
    for (int j = 2; j <= len; j++) {
      int x = s[j - 1] - 'a', y = s[j] - 'a';
      if ((to[x] != -1 && to[x] != y) || (from[y] != -1 && from[y] != x)) {
        flag = false;
        break;
      }
      vis[x] = vis[y] = 1;
      to[x] = y;
      from[y] = x;
    }
    vis[s[1] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (from[i] == -1 && !vis2[i] && vis[i])
      for (int j = i; j != -1; j = to[j]) {
        ans[++cnt] = j;
        vis2[j] = 1;
      }
  for (int i = 0; i < 26; i++)
    if (vis[i] && !vis2[i]) flag = false;
  if (!flag)
    puts("NO");
  else
    for (int i = 1; i <= cnt; i++) printf("%c", 'a' + ans[i]);
  return 0;
}
