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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 100005;
string s[N];
string ans;
int cnt[26];
set<int> inG[26], outG[26];
int vis[26];
void dfs(int u) {
  vis[u] = 1;
  ans.push_back('a' + u);
  for (auto it = outG[u].begin(); it != outG[u].end(); it++) {
    int v = *it;
    if (!vis[v]) dfs(v);
  }
}
int occur[26];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int ok = 1;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    memset(cnt, 0, sizeof(cnt));
    for (int j = 0; j < (int)(s[i].size()); j++) {
      cnt[s[i][j] - 'a']++;
      occur[s[i][j] - 'a'] = 1;
    }
    for (int i = 0; i < 26; i++)
      if (cnt[i] >= 2) ok = 0;
  }
  if (!ok) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)(s[i].size()) - 1; j++) {
      outG[s[i][j] - 'a'].insert(s[i][j + 1] - 'a');
      inG[s[i][j + 1] - 'a'].insert(s[i][j] - 'a');
    }
  }
  for (int i = 0; i < 26; i++) {
    if ((int)(inG[i].size()) >= 2 || (int)(outG[i].size()) >= 2) {
      ok = 0;
    }
  }
  if (!ok) {
    cout << "NO\n";
    return 0;
  }
  ans = "";
  for (int i = 0; i < 26; i++) {
    if ((int)(inG[i].size()) == 0 && occur[i] == 1) dfs(i);
  }
  for (int i = 0; i < 26; i++)
    if (occur[i] == 1 && vis[i] == 0) {
      cout << "NO\n";
      return 0;
    }
  cout << ans << "\n";
  return 0;
}
