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;
string s[100005];
int n;
int pre[400], after[400], dd[400];
int main() {
  memset(dd, 0, sizeof(dd));
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> s[i];
  for (int i = 'a'; i <= 'z'; i++) pre[i] = -1, after[i] = -1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < s[i].size(); j++) dd[s[i][j]] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < s[i].size() - 1; j++) {
      char x = s[i][j];
      char y = s[i][j + 1];
      if (after[x] != -1 && pre[y] != x) return cout << "NO", 0;
      if (pre[y] != -1 && pre[y] != x) return cout << "NO", 0;
      after[x] = y;
      pre[y] = x;
    }
  }
  for (int i = 'a'; i <= 'z'; i++)
    if (dd[i] != 0) {
      int j = i;
      while (1) {
        j = after[j];
        if (j == -1) break;
        if (j == i) return cout << "NO", 0;
      }
    }
  for (int i = 'a'; i <= 'z'; i++)
    if (dd[i] != 0 && pre[i] == -1) {
      int j = i;
      cout << (char)i;
      while (1) {
        j = after[j];
        if (j == -1) break;
        cout << (char)j;
      }
    }
  return 0;
}
