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 ln[35][35];
int p[35], vis[35];
int in[35], out[35];
string s[100005], ans;
int n;
int dfs(int now) {
  vis[now] = 1;
  ans = ans + (char)(now + 'a');
  for (int j = 0; j < 26; j++) {
    if (ln[now][j]) {
      if (vis[j]) return 0;
      return dfs(j);
    }
  }
  return 1;
}
int main() {
  int len, flag = 0;
  scanf("%d", &n);
  memset(ln, 0, sizeof(ln));
  memset(p, 0, sizeof(p));
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    len = s[i].length();
    for (int j = 0; j < len - 1; j++) ln[s[i][j] - 'a'][s[i][j + 1] - 'a'] = 1;
    if (len == 1) p[s[i][0] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++) {
    if (ln[i][i]) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 0; i < 26; i++)
    for (int j = 0; j < 26; j++) {
      if (ln[i][j]) {
        in[j]++;
        out[i]++;
      }
    }
  for (int i = 0; i < 26; i++) {
    if (in[i] > 1 || out[i] > 1) {
      printf("NO\n");
      return 0;
    }
  }
  memset(vis, 0, sizeof(vis));
  ans = "";
  for (int i = 0; i < 26; i++) {
    if (out[i] != 0 && in[i] == 0) {
      if (!dfs(i)) {
        printf("NO\n");
        return 0;
      }
    }
    if (p[i] && in[i] == 0 && out[i] == 0) ans = ans + (char)(i + 'a');
  }
  for (int i = 0; i < 26; i++) {
    if (in[i] != 0 || out[i] != 0) {
      if (!vis[i]) {
        printf("NO\n");
        return 0;
      }
    }
  }
  cout << ans << endl;
  return 0;
}
