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 adj[30][30], in[30], out[30];
bool vis[30], need[30], cycle;
string chain;
void dfs(int u) {
  if (vis[u]) {
    cycle = 1;
    return;
  }
  vis[u] = 1;
  chain += char('a' + u);
  for (int v = 0; v < (int)(30); v++)
    if (adj[u][v]) dfs(v);
}
void no() {
  cout << "NO\n";
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  while (n--) {
    string s;
    cin >> s;
    for (char c : s) need[c - 'a'] = 1;
    for (int i = 0; i < (int)(((int)(s).size()) - 1); i++) {
      int u = s[i] - 'a', v = s[i + 1] - 'a';
      if (!adj[u][v]) {
        adj[u][v]++;
        out[u]++;
        in[v]++;
      }
    }
  }
  bool ok = 1;
  for (int u = 0; u < (int)(30); u++) ok &= (in[u] <= 1 && out[u] <= 1);
  if (!ok) no();
  vector<string> ans;
  for (int u = 0; u < (int)(30); u++)
    if (!in[u] && need[u]) {
      chain = "";
      cycle = 0;
      dfs(u);
      ans.push_back(chain);
    }
  for (int u = 0; u < (int)(30); u++)
    if (!vis[u] && need[u]) ok = 0;
  if (!ok) no();
  sort((ans).begin(), (ans).end());
  string res;
  for (auto &x : ans) res += x;
  cout << res << endl;
  return 0;
}
