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 ind = 0;
int n;
char nekst[300];
char buff[100007];
bool vis[300];
int cnt[300];
int deg[300];
set<char> S;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf(" %s", buff);
    int j;
    for (j = 0; buff[j + 1]; j++) {
      char a = buff[j];
      char b = buff[j + 1];
      cnt[a]++;
      if (nekst[a] != 0 && nekst[a] != b) {
        printf("NO");
        return 0;
      }
      nekst[a] = b;
      deg[b]++;
    }
    cnt[buff[j]]++;
  }
  int all = 0;
  for (char i = 'a'; i <= 'z'; i++) {
    if (cnt[i] && !deg[i]) S.insert(i);
    if (cnt[i]) all++;
  }
  int used = 0;
  while (!S.empty()) {
    char c = *S.begin();
    S.erase(S.begin());
    while (c && !vis[c]) {
      vis[c] = true;
      buff[ind++] = c;
      used++;
      c = nekst[c];
    }
    if (c && vis[c]) {
      printf("NO");
      return 0;
    }
  }
  buff[ind++] = 0;
  if (used < all) {
    printf("NO");
    return 0;
  }
  printf("%s", buff);
  return 0;
}
