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 double pi = acos(-1);
const int oo = 1 << 30;
const long long OO = 1e18;
const int N = 2e5 + 100;
int gi() {
  int w = 0;
  bool q = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') q = 0, c = getchar();
  while (c >= '0' && c <= '9') w = w * 10 + c - '0', c = getchar();
  return q ? w : -w;
}
int f[N], g[N];
bool o[N];
string ans;
int sum;
inline void dfs(int k) {
  ans.push_back(k - 1 + 'a');
  sum++;
  if (f[k]) dfs(f[k]);
}
int main() {
  int n = gi(), i;
  string x;
  while (n--) {
    cin >> x;
    if (x.size() == 1)
      o[x[0] - 'a' + 1] = true;
    else
      for (i = 1; i < x.size(); i++) {
        if (!f[x[i - 1] - 'a' + 1])
          f[x[i - 1] - 'a' + 1] = x[i] - 'a' + 1;
        else if (f[x[i - 1] - 'a' + 1] != x[i] - 'a' + 1)
          return puts("NO"), 0;
        if (!g[x[i] - 'a' + 1])
          g[x[i] - 'a' + 1] = x[i - 1] - 'a' + 1;
        else if (g[x[i] - 'a' + 1] != x[i - 1] - 'a' + 1)
          return puts("NO"), 0;
      }
  }
  for (i = 1; i <= 26; i++) {
    if (!g[i] && (o[i] || f[i]))
      dfs(i);
    else
      sum += !g[i] && !f[i] && !o[i];
  }
  if (sum != 26)
    puts("NO");
  else
    cout << ans << endl;
  return 0;
}
