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 MAXN = 26 + 10;
set<int> ne[MAXN];
bool must[MAXN], mark[MAXN];
vector<int> ve;
string s;
int n, dig[MAXN];
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s;
    set<char> uni;
    for (auto j : s) {
      uni.insert(j);
      must[j - 'a'] = true;
    }
    if (uni.size() != s.size()) {
      return cout << "NO" << endl, 0;
    }
    for (int j = 1; j < s.size(); j++) {
      dig[s[j] - 'a']++;
      ne[s[j - 1] - 'a'].insert(s[j] - 'a');
    }
  }
  for (int i = 0; i < 26; i++) {
    if (ne[i].size() > 1) return cout << "NO" << endl, 0;
    if (!dig[i] && must[i]) ve.push_back(i);
  }
  s = "";
  for (auto i : ve) {
    int me = i;
    while (~me) {
      if (mark[me]) {
        return cout << "NO" << endl, 0;
      }
      mark[me] = true;
      s += me + 'a';
      if (ne[me].size())
        me = *ne[me].begin();
      else
        me = -1;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (must[i] && !mark[i]) return cout << "NO" << endl, 0;
  }
  cout << s << endl;
  return 0;
}
