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;
char ch[100005];
int mp[27][27];
bool f[27];
bool vis[27];
char ans[27];
int main() {
  int n;
  scanf("%d", &n);
  int len;
  int u, v;
  memset(f, 0, sizeof(f));
  while (n--) {
    scanf("%s", ch);
    len = strlen(ch);
    vis[ch[0] - 'a'] = 1;
    for (int i = 1; i < len; i++) {
      u = ch[i - 1] - 'a';
      v = ch[i] - 'a';
      mp[u][v] = 1;
      f[v] = 1;
      vis[v] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    int x = 0, y = 0;
    for (int j = 0; j < 26; j++) {
      if (mp[i][j]) x++;
      if (mp[j][i]) y++;
    }
    if (x > 1 || y > 1) {
      puts("NO");
      return 0;
    }
  }
  int cou = 0;
  for (int i = 0; i < 26; i++)
    if (!f[i] && vis[i]) {
      int k = i;
      while (k != -1) {
        ans[cou++] = 'a' + k;
        int t = -1;
        for (int j = 0; j < 26; j++)
          if (mp[k][j]) t = j;
        k = t;
      }
    }
  int tmp = 0;
  for (int i = 0; i < 26; i++)
    if (vis[i]) tmp++;
  if (cou < tmp)
    puts("NO");
  else {
    for (int i = 0; i < cou; i++) putchar(ans[i]);
    puts("");
  }
  return 0;
}
