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;
vector<set<long long>> g(26);
vector<long long> in(26, 0);
vector<long long> used(26, 0);
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  while (n--) {
    string s;
    cin >> s;
    for (long long i = 1; i < s.length(); i++) {
      if (g[s[i - 1] - 'a'].find(s[i] - 'a') == g[s[i - 1] - 'a'].end())
        in[s[i] - 'a']++;
      g[s[i - 1] - 'a'].insert(s[i] - 'a');
    }
    for (char c : s) used[c - 'a'] = 1;
  }
  queue<long long> q;
  for (long long i = 0; i < 26; i++)
    if (!in[i] && used[i])
      q.push(i);
    else if (in[i] > 1) {
      cout << "NO";
      return 0;
    }
  vector<long long> usedd(26, 0);
  string ans = "";
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    bool done = false;
    ans = ans + (char)(cur + 'a');
    while (!done) {
      usedd[cur] = 1;
      if (g[cur].size() == 1 && !usedd[*g[cur].begin()]) {
        cur = *g[cur].begin();
        ans = ans + (char)(cur + 'a');
      } else if (g[cur].size() == 0)
        done = true;
      else {
        cout << "NO";
        return 0;
      }
    }
  }
  for (long long i = 0; i < 26; i++)
    if (!usedd[i] && used[i]) {
      cout << "NO";
      return 0;
    }
  cout << (ans.length() ? ans : "NO");
  return 0;
}
