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 N = 2e5 + 10;
const int md = 998244353;
const long long INF = 4e18;
int n;
string s;
int bef[30];
int have[30];
int used[30];
int yes[30];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s;
    have[s[0] - 'a'] = 1;
    for (int j = 0; j < 30; j++) used[j] = 0;
    used[s[0] - 'a'] = 1;
    for (int j = 1; j < s.size(); j++) {
      if (!bef[s[j - 1] - 'a']) bef[s[j - 1] - 'a'] = s[j];
      if (s[j] != bef[s[j - 1] - 'a']) return cout << "NO", 0;
      have[s[j] - 'a'] = 1;
      if (used[s[j] - 'a'] == 1) return cout << "NO", 0;
      used[s[j] - 'a'] = 1;
      yes[s[j] - 'a'] = 1;
    }
  }
  for (int i = 0; i < 30; i++) used[i] = 0;
  string k;
  for (int i = 'a'; i <= 'z'; i++) {
    if (have[i - 'a']) {
      if (used[i - 'a']) continue;
      if (yes[i - 'a']) continue;
      k += i;
      used[i - 'a'] = 1;
      int c = i;
      while (bef[c - 'a']) {
        c = bef[c - 'a'];
        if (used[c - 'a']) return cout << "NO", 0;
        used[c - 'a'] = 1;
        k += c;
      }
    }
  }
  if (!int(k.size())) return cout << "NO", 0;
  for (int i = 'a'; i <= 'z'; i++) {
    if (have[i - 'a']) {
      for (int j = 0; j < int(k.size()); j++) {
        if (k[j] == i) break;
        if (j == int(k.size()) - 1) return cout << "NO", 0;
      }
    }
  }
  cout << k;
  return 0;
}
