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;
int n;
map<char, vector<char> > e;
map<char, int> cl;
map<char, bool> was;
map<char, bool> bukv;
map<char, int> c1;
vector<char> d;
string ans = "";
map<pair<char, char>, bool> m;
bool dfs(char v) {
  cl[v] = 1;
  for (auto to : e[v]) {
    if (cl[to] == 0) {
      if (dfs(to)) return true;
    } else if (cl[to] == 1) {
      return true;
    }
  }
  cl[v] = 2;
  return false;
}
bool dfs2(char v) {
  was[v] = 1;
  ans += v;
  for (auto to : e[v]) {
    dfs2(to);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    d.push_back(s[0]);
    for (int i = 1; i < s.size(); i++) {
      if (!m[{s[i - 1], s[i]}]) {
        e[s[i - 1]].push_back(s[i]);
        c1[s[i]]++;
      }
      bukv[s[i]] = 1;
      m[{s[i - 1], s[i]}] = 1;
      if (e[s[i - 1]].size() > 1) {
        cout << "NO";
        return 0;
      }
      if (c1[s[i]] > 1) {
        cout << "NO";
        return 0;
      }
    }
  }
  for (char c = 'a'; c <= 'z'; c++) {
    if (!cl[c]) {
      if (dfs(c)) {
        cout << "NO";
        return 0;
      }
    }
  }
  sort(d.begin(), d.end());
  for (auto i : d) {
    if (!was[i] && !bukv[i]) dfs2(i);
  }
  cout << ans;
  return 0;
}
