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 = 30;
bitset<N> vis;
vector<int> graph[N];
vector<string> res;
string s;
int n, sz, a, b, cnt, usd, parent[N];
inline string dfs(int cur) {
  string res = "";
  usd++;
  for (int it : graph[cur]) return ((char)(cur + 'a')) + dfs(it);
  return ((char)(cur + 'a')) + res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(parent, -1, sizeof(parent));
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s;
    sz = s.size();
    for (int j = 0; j < sz; j++) {
      a = s[j] - 'a';
      if (!vis[a]) {
        cnt++;
        vis[a] = 1;
      }
      if (j + 1 < sz) {
        b = s[j + 1] - 'a';
        for (int it : graph[a])
          if (it != b || parent[it] != a) {
            cout << "NO\n";
            return 0;
          }
        graph[a].push_back(b);
        if (parent[b] != -1 && parent[b] != a) {
          cout << "NO\n";
          return 0;
        }
        parent[b] = a;
      }
    }
  }
  for (char ch = 'a'; ch <= 'z'; ch++) {
    a = ch - 'a';
    if (vis[a] && parent[a] == -1) {
      res.push_back(dfs(a));
    }
  }
  if (usd != cnt) {
    cout << "NO\n";
    return 0;
  }
  for (string st : res) cout << st;
  cout << "\n";
  return 0;
}
