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 N = 1e6 + 5;
string s, ans;
int n, b[505], us[505], used[505];
vector<int> g[505];
int gg[505][505];
const int mod = 1e9 + 7;
void dfs(int v) {
  used[v] = 1;
  if (g[v].size() > 1) {
    cout << "NO";
    exit(0);
  }
  for (int i = 0; i < g[v].size(); i++) {
    if (!used[g[v][i]]) {
      dfs(g[v][i]);
    } else if (used[g[v][i]]) {
      cout << "NO";
      exit(0);
    }
  }
  ans += v;
  used[v] = 2;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    for (int j = 0; j < s.size() - 1; j++) {
      us[s[j]] = 1;
      if (!gg[s[j]][s[j + 1]]) g[s[j]].push_back(s[j + 1]);
      gg[s[j]][s[j + 1]] = 1;
      b[s[j + 1]] = 1;
    }
    us[s[s.size() - 1]] = 1;
  }
  for (int i = 'z'; i >= 'a'; i--) {
    if (us[i]) {
      if (!b[i] && !used[i]) {
        dfs(i);
      }
    }
  }
  for (int i = 'a'; i <= 'z'; i++) {
    if (us[i] && !used[i]) {
      cout << "NO";
      return 0;
    }
  }
  if (ans.size() == 0) {
    cout << "NO";
    return 0;
  }
  reverse(ans.begin(), ans.end());
  cout << ans;
}
