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;
int a, s, d[28], f[28], g[28], h, j, k, l, i, n, m;
string pas;
void dfs(int idx) {
  pas += (char)(idx + 96);
  h++;
  if (!d[idx]) return;
  dfs(d[idx]);
}
string x;
int main() {
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    if (x.size() == 1) g[x[0] - 96] = 1;
    for (a = 0; a < x.size() - 1; a++) {
      if (d[x[a] - 96] && d[x[a] - 96] != x[a + 1] - 96) {
        cout << "NO";
        return 0;
      }
      if (f[x[a + 1] - 96] && f[x[a + 1] - 96] != x[a] - 96) {
        cout << "NO";
        return 0;
      }
      d[x[a] - 96] = x[a + 1] - 96;
      f[x[a + 1] - 96] = x[a] - 96;
    }
  }
  for (i = 1; i <= 26; i++) {
    if (!f[i] && (d[i] || g[i])) dfs(i);
    if (!d[i] && !f[i] && !g[i]) h++;
  }
  if (h != 26)
    cout << "NO";
  else
    cout << pas;
  return 0;
}
