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 N = 100005;
map<char, long long> m;
map<char, char> g;
map<char, long long> vis;
void solve() {
  long long n;
  string s;
  cin >> n;
  long long c = 0;
  set<char> second;
  for (char c = 'a'; c <= 'z'; c = c + 1) {
    second.insert(c);
  }
  while (n--) {
    cin >> s;
    for (long long i = 0; i < s.size(); i++) {
      if (m[s[i]] == 0) {
        m[s[i]] = 1;
        c++;
      }
      if (i > 0) {
        if (g.count(s[i - 1]) and g[s[i - 1]] != s[i]) {
          cout << "NO" << '\n';
          return;
        }
        g[s[i - 1]] = s[i];
        if (second.find(s[i]) != second.end()) second.erase(s[i]);
      }
    }
  }
  priority_queue<char, vector<char>, greater<char> > pq;
  for (char c = 'a'; c <= 'z'; c = c + 1) {
    if (m[c] == 1 and second.find(c) != second.end()) pq.push(c);
  }
  string ans;
  while (!pq.empty()) {
    char cp = pq.top();
    pq.pop();
    while (g.count(cp)) {
      ans.push_back(cp);
      c--;
      cp = g[cp];
      if (vis.count(cp)) {
        cout << "NO" << '\n';
        return;
      }
      vis[cp] = 1;
    }
    ans.push_back(cp);
    c--;
  }
  if (c != 0)
    cout << "NO" << '\n';
  else
    cout << ans << '\n';
}
signed main() {
  long long t;
  ios_base::sync_with_stdio(false);
  t = 1;
  while (t--) solve();
  return 0;
}
