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 long long M = 1e6 + 6;
const long long INF = 1e18 + 5;
const long long MOD = 1e9 + 7;
const long long MAX = 25;
const long long N = 1e7 + 5;
const double PI = 3.14159265358979323846264338327950288419716939937510;
long long n, to[30], from[30], vis[30], mark[30];
string str;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  memset(to, -1, sizeof to);
  memset(from, -1, sizeof from);
  for (long long z = 1; z <= n; z++) {
    string s;
    cin >> s;
    for (long long i = 1; i <= s.length() - 1; i++) {
      long long a = s[i - 1] - 'a';
      long long b = s[i] - 'a';
      if ((to[a] != -1 && to[a] != b) || (from[b] != -1 && from[b] != a)) {
        cout << "NO\n";
        return 0;
      }
      to[a] = b;
      from[b] = a;
      vis[b] = 1;
    }
    vis[s[0] - 'a'] = 1;
  }
  for (long long i = 0; i < 26; i++) {
    if (vis[i] == 1 && from[i] == -1) {
      for (long long j = i; j != -1; j = to[j]) {
        mark[j] = 1;
        str += (char)('a' + j);
      }
    }
  }
  for (long long i = 0; i < 26; i++) {
    if (vis[i] == 1 && mark[i] == 0) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << str << "\n";
}
