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 long long int c = 26;
vector<string> s;
vector<long long int> mp[c];
bitset<c> vis, used;
long long int n, deg[26];
string ans;
bool topological() {
  long long int cnt = c;
  deque<long long int> q;
  for (int i = 0; i < c; i++) {
    if (!deg[i] && used[i])
      q.push_back(i);
    else if (!used[i])
      cnt--;
  }
  while (q.size()) {
    long long int cur = q.front();
    q.pop_front();
    ans += char(cur + 'a'), vis[cur] = true, cnt--;
    for (long long int i : mp[cur]) {
      if (vis[i]) continue;
      deg[i]--;
      if (!deg[i]) {
        q.push_front(i);
      }
    }
  }
  if (cnt) return false;
  return true;
}
bool check(string s) {
  bool succeed = false;
  if (s.length() > ans.length()) return false;
  for (int i = 0; i < ans.length() - s.length() + 1; i++) {
    succeed = false;
    for (int j = 0; j < s.length(); j++) {
      if (s[j] != ans[j + i]) break;
      if (j == s.length() - 1) succeed = true;
    }
    if (succeed) return true;
  }
  return false;
}
bool solve() {
  for (string ss : s) {
    if (ss.length() > c) return false;
    for (int i = 0; i < ss.length() - 1; i++) {
      int f = ss[i] - 'a', to = ss[i + 1] - 'a';
      mp[f].push_back(to), deg[to]++;
      used[f] = true, used[to] = true;
    }
    int lst = ss[ss.length() - 1] - 'a';
    used[lst] = true;
  }
  for (int i = 0; i < c; i++) {
    sort(mp[i].begin(), mp[i].end());
  }
  if (!topological()) return false;
  for (string ss : s) {
    if (!check(ss)) return false;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n, s.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
  }
  if (solve()) {
    cout << ans << "\n";
  } else
    cout << "NO\n";
  return 0;
}
