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 T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
const long long N = 2e5 + 5;
long long n, present[N];
string ans = "";
string a[N];
set<long long> g[N], rg[N];
void dfs(long long k) {
  present[k] = 0;
  ans += (char)(k + 'a');
  for (auto it : g[k]) dfs(it);
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    string s = a[i];
    long long sz = s.size();
    for (long long j = 0; j < sz - 1; j++) {
      g[s[j] - 'a'].insert(s[j + 1] - 'a');
      rg[s[j + 1] - 'a'].insert(s[j] - 'a');
    }
    for (long long j = 0; j < sz; j++) {
      present[s[j] - 'a'] = 1;
    }
  }
  for (long long i = 0; i < 26; i++) {
    if (g[i].size() > 1 || rg[i].size() > 1) {
      cout << "NO";
      return 0;
    }
  }
  for (long long i = 0; i < 26; i++) {
    if (rg[i].size() == 0 && present[i]) dfs(i);
  }
  for (long long i = 0; i < 26; i++) {
    if (present[i]) {
      cout << "NO";
      return 0;
    }
  }
  cout << ans;
  return 0;
}
