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>
const double Pi = acos(-1.0);
using namespace std;
int n;
char s[100005];
bool in[30];
int pre[30], nxt[30];
bool vis[30];
int main(int argc, char** argv) {
  scanf("%d", &n);
  for (int i = 0; i < 26; i++) pre[i] = nxt[i] = -1;
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    in[s[0] - 'a'] = true;
    for (int j = 1; j < (int)strlen(s); j++) {
      int l1 = s[j] - 'a', l2 = s[j - 1] - 'a';
      in[l1] = in[l2] = true;
      if (l1 == l2 || (nxt[l2] != -1 && nxt[l2] != l1) ||
          (pre[l1] != -1 && pre[l1] != l2)) {
        puts("NO");
        return 0;
      }
      pre[l1] = l2;
      nxt[l2] = l1;
    }
  }
  string ans = "";
  int cnt = 0;
  for (int i = 0; i < 26; i++) {
    if (in[i]) {
      cnt++;
      if (pre[i] == -1) {
        for (int j = i; j != -1; j = nxt[j]) {
          if (vis[j]) {
            puts("NO");
            return 0;
          }
          vis[j] = true;
          ans += j + 'a';
        }
      }
    }
  }
  puts(((int)ans.size() == cnt) ? ans.c_str() : "NO");
  return 0;
}
