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 MAXN = (int)1e5;
string str[MAXN + 1];
inline void fail() {
  cout << "NO";
  exit(0);
}
int fr[26], nxt[26];
bool first[26], in[26];
bool vis[26];
string sol[MAXN + 1];
int main() {
  int i, j, n;
  ios::sync_with_stdio(false);
  cin >> n;
  memset(nxt, -1, sizeof(nxt));
  for (i = 1; i <= n; i++) {
    cin >> str[i];
    memset(fr, 0, sizeof(fr));
    for (j = 0; j < str[i].size(); j++) {
      str[i][j] -= 'a';
    }
    first[str[i][0]] = 1;
    for (j = 0; j + 1 < str[i].size(); j++) {
      in[str[i][j + 1]] = 1;
      if (nxt[str[i][j]] != -1 && nxt[str[i][j]] != str[i][j + 1]) fail();
      nxt[str[i][j]] = str[i][j + 1];
      fr[str[i][j]]++;
      if (fr[str[i][j]] == 2) fail();
    }
  }
  int cnt = 0;
  for (i = 0; i < 26; i++) {
    if (first[i] == 1 && vis[i] == 0 && in[i] == 0) {
      j = i;
      cnt++;
      while (j > -1 && vis[j] == 0) {
        vis[j] = 1;
        sol[cnt].push_back(j + 'a');
        j = nxt[j];
      }
      if (j > -1) fail();
    }
  }
  for (i = 0; i < 26; i++) {
    if (first[i] == 1 && vis[i] == 0) fail();
  }
  sort(sol + 1, sol + cnt + 1);
  for (i = 1; i <= cnt; i++) {
    cout << sol[i];
  }
  return 0;
}
