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 n;
vector<int> before[30];
vector<int> after[30];
map<int, bool> exists;
bool used[30];
int main() {
  scanf("%d", &n);
  string str;
  for (int i = 0; i < n; i++) {
    cin >> str;
    for (int j = 0; j < str.size(); j++) {
      exists[str[j] - 'a'] = true;
      int v = str[j] - 'a';
      if (j != 0) {
        int u = str[j - 1] - 'a';
        if (before[v].size() > 0 && before[v][0] != u) {
          printf("NO");
          return 0;
        }
        if (before[v].size() == 0) before[v].push_back(u);
      }
      if (j != str.size() - 1) {
        int u = str[j + 1] - 'a';
        if (after[v].size() > 0 && after[v][0] != u) {
          printf("NO");
          return 0;
        }
        if (after[v].size() == 0) after[v].push_back(u);
      }
    }
  }
  str = "";
  vector<int> start;
  for (int i = 0; i < 26; i++) {
    if (before[i].size() == 0 && exists[i]) start.push_back(i);
  }
  if (start.size() == 0) {
    printf("NO");
    return 0;
  }
  for (int i = 0; i < start.size(); i++) {
    int v = start[i];
    while (true) {
      str += ('a' + v);
      used[v] = true;
      if (after[v].size() == 0) break;
      v = after[v][0];
    }
  }
  for (int i = 0; i < 26; i++) {
    if (exists[i] && !used[i]) {
      printf("NO");
      return 0;
    }
  }
  cout << str;
  return 0;
}
