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;
map<char, char> bef, aft;
set<char> need;
int main() {
  ios_base::sync_with_stdio(false);
  for (char c = 'a'; c <= 'z'; c++) {
    bef[c] = c;
    aft[c] = c;
  }
  int n;
  cin >> n;
  bool ok = true;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (char c : s) {
      need.insert(c);
    }
    string ss = s;
    sort(ss.begin(), ss.end());
    for (int j = 1; j < ss.size(); j++) {
      if (ss[j] == ss[j - 1]) ok = false;
    }
    for (int j = 1; j < s.size(); j++) {
      if (!(aft[s[j - 1]] == s[j - 1] || aft[s[j - 1]] == s[j])) {
        ok = false;
      }
      if (!(bef[s[j]] == s[j - 1] || bef[s[j]] == s[j])) {
        ok = false;
      } else {
        aft[s[j - 1]] = s[j];
        bef[s[j]] = s[j - 1];
      }
    }
  }
  vector<string> red;
  int db = 0;
  for (char c : need) {
    if (bef[c] != c) {
      continue;
    }
    db++;
    string d(1, c);
    while (aft[c] != c) {
      c = aft[c];
      d.push_back(c);
      db++;
    }
    red.push_back(d);
  }
  sort(red.begin(), red.end());
  if (!ok || db != need.size()) {
    cout << "NO" << endl;
  } else {
    for (string f : red) {
      cout << f;
    }
    cout << endl;
  }
  return 0;
}
