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 int64_t INF = 1e18 + 10;
const int N = 2e7 + 10;
int MOD = 1000000007;
int g[26], in[26], used[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 26; i++) g[i] = -1;
  int n;
  cin >> n;
  vector<string> a(n, "");
  bool arr = 1;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (auto j : s) used[j - 'a'] = 1;
    for (int j = 1; j < s.size(); j++) {
      int u = s[j - 1] - 'a';
      if (g[u] != -1 && g[u] != s[j] - 'a' || s[j] == s[j - 1]) arr = 0;
      g[s[j - 1] - 'a'] = s[j] - 'a';
      in[s[j] - 'a']++;
    }
  }
  if (!arr) {
    cout << "NO";
    return 0;
  }
  string ans = "";
  for (int i = 0; i < 26; i++) {
    if (in[i] == 0 && used[i]) {
      int x = i;
      while (x != -1) {
        used[x]--;
        if (used[x] < 0) break;
        ans += 'a' + x;
        x = g[x];
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (used[i] != 0) arr = 0;
  }
  if (arr)
    cout << ans;
  else
    cout << "NO";
}
