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 = 200050;
long long inp1, inp2, inp3;
long long n, fr[MAXN], to[MAXN], vis[MAXN];
char s[MAXN];
string ans;
int main() {
  memset(fr, -1, sizeof fr);
  memset(to, -1, sizeof to);
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%s", s);
    long long len = strlen(s);
    for (long long i = 0; i < len; i++) {
      vis[s[i] - 'a'] = 1;
      if (i > 0) {
        if (~fr[s[i] - 'a'] && fr[s[i] - 'a'] != s[i - 1] - 'a') {
          printf("NO");
          return 0;
        }
        fr[s[i] - 'a'] = s[i - 1] - 'a';
      }
      if (i < len - 1) {
        if (~to[s[i] - 'a'] && to[s[i] - 'a'] != s[i + 1] - 'a') {
          printf("NO");
          return 0;
        }
        to[s[i] - 'a'] = s[i + 1] - 'a';
      }
    }
  }
  for (long long i = 0; i < 27; i++) {
    if (vis[i] && fr[i] == -1) {
      long long tmp = i;
      while (~tmp) {
        vis[tmp] = 0;
        ans += (char)(tmp + 'a');
        tmp = to[tmp];
      }
    }
  }
  for (long long i = 0; i < 27; i++) {
    if (vis[i]) {
      printf("NO");
      return 0;
    }
  }
  cout << ans;
  return 0;
}
