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<int> g[26];
vector<string> ans;
bool visited[26] = {};
int cnt = 0;
int n;
string a[100005];
bool dfs(int u) {
  visited[u] = 1;
  ans[cnt].push_back('a' + u);
  if (g[u].size() > 1) {
    return 0;
  }
  if (g[u].empty()) {
    return 1;
  }
  int v = g[u][0];
  if (visited[v]) {
    return 0;
  }
  return dfs(v);
}
bool check(string& fin) {
  for (int i = 0; i < n; i++) {
    if (fin.find(a[i]) == std::string::npos) {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> n;
  vector<int> par(26, -1);
  vector<int> ispres(26, 0);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    for (int j = 0; j < a[i].size() - 1; j++) {
      ispres[a[i][j] - 'a'] = 1;
      g[a[i][j] - 'a'].push_back(a[i][j + 1] - 'a');
      par[a[i][j + 1] - 'a'] = a[i][j] - 'a';
    }
    ispres[a[i].back() - 'a'] = 1;
  }
  vector<int>::iterator ip;
  for (int i = 0; i < 26; i++) {
    sort(g[i].begin(), g[i].end());
    ip = unique(g[i].begin(), g[i].end());
    g[i].resize(distance(g[i].begin(), ip));
  }
  for (int i = 0; i < 26; i++) {
    if (par[i] == -1 && ispres[i]) {
      ans.emplace_back();
      if (!dfs(i)) {
        cout << "NO" << endl;
        return 0;
      }
      cnt++;
    }
  }
  string fin;
  for (int i = 0; i < ans.size(); i++) {
    fin += ans[i];
  }
  if (!check(fin)) {
    cout << "NO" << endl;
    return 0;
  }
  cout << fin << endl;
  return 0;
}
