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 + 9;
int n, cnt;
bool taken[N];
string s;
vector<char> ans;
bool occ[28], cov[28];
int main() {
  cin >> n;
  char a[28], p[28];
  for (int i = 0; i < 26; i++) {
    a[i] = p[i] = 'A';
  }
  for (int i = 0; i < n; i++) {
    cin >> s;
    for (int j = 0; j < (int)s.size() - 1; j++) {
      occ[s[j] - 'a'] = 1;
      if (a[s[j] - 'a'] == s[j + 1] || a[s[j] - 'a'] == 'A') {
        a[s[j] - 'a'] = s[j + 1];
      } else {
        cout << "NO\n";
        return 0;
      }
    }
    occ[s[(int)s.size() - 1] - 'a'] = 1;
    for (int j = 1; j < (int)s.size(); j++) {
      if (p[s[j] - 'a'] != 'A' && p[s[j] - 'a'] != s[j - 1]) {
        cout << "NO\n";
        return 0;
      }
      p[s[j] - 'a'] = s[j - 1];
    }
  }
  set<char> ss;
  for (int i = 0; i < 26; i++) {
    if (occ[i]) ss.insert(i + 'a');
  }
  for (int i = 0; i < 26; i++) {
    if (a[i] != 'A' && ss.count(a[i])) {
      ss.erase(a[i]);
    }
  }
  for (auto it : ss) {
    char j = it;
    while (j != 'A') {
      ans.push_back(j);
      occ[j - 'a'] = 0;
      j = a[j - 'a'];
    }
  }
  for (int i = 0; i < 26; i++) {
    if (occ[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  for (auto it : ans) cout << it;
  cout << endl;
  return 0;
}
