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 MM = 2e5 + 5, MOD = 1e9 + 7;
int N, c[26], h[26], z[26], v[26], in[26];
string s, ans;
vector<int> adj[MM];
void dfs(int u) {
  v[u] = 1;
  ans += (char('a' + u));
  if (adj[u].size() > 1) {
    cout << "NO"
         << "\n";
    exit(0);
  }
  for (int x : adj[u]) {
    if (!v[x])
      dfs(x);
    else {
      cout << "NO"
           << "\n";
      exit(0);
    }
  }
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  cin >> N;
  memset(c, -1, sizeof(c));
  for (int i = 1; i <= N; i++) {
    cin >> s;
    memset(h, 0, sizeof(h));
    h[s[0] - 'a'] = 1;
    z[s[0] - 'a'] = 1;
    for (int j = 1; j < s.size(); j++) {
      if (s[j] == s[j - 1] ||
          (c[s[j - 1] - 'a'] != -1 && c[s[j - 1] - 'a'] != (s[j] - 'a')) ||
          h[s[j] - 'a']) {
        cout << "NO"
             << "\n";
        return 0;
      }
      if (c[s[j - 1] - 'a'] == -1) {
        c[s[j - 1] - 'a'] = s[j] - 'a';
        in[s[j] - 'a']++;
        adj[s[j - 1] - 'a'].push_back(s[j] - 'a');
      }
      h[s[j] - 'a'] = 1;
      z[s[j] - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++)
    if (z[i] && !v[i] && in[i] == 0) dfs(i);
  for (int i = 0; i < 26; i++) {
    if (z[i] && !v[i]) {
      cout << "NO"
           << "\n";
      exit(0);
    }
  }
  cout << ans << "\n";
}
