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 N = 100010;
int n, pre[26], nxt[26], can[26];
char s[N];
int main() {
  scanf("%d", &n);
  int all = 0;
  memset(pre, -1, sizeof(pre));
  memset(nxt, -1, sizeof(nxt));
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    int frq = 0;
    int l = strlen(s);
    for (int i = 0; i < l; i++) {
      int idx = s[i] - 'a';
      if (((frq >> idx) & 1) == 1) {
        puts("NO");
        return 0;
      }
      frq += (1 << idx);
      if (can[idx] == false) all++;
      can[idx] = true;
      if (i > 0) {
        if (pre[idx] != -1 && pre[idx] != s[i - 1] - 'a') {
          puts("NO");
          return 0;
        }
        pre[idx] = s[i - 1] - 'a';
      }
      if (i < l - 1) {
        if (nxt[idx] != -1 && nxt[idx] != s[i + 1] - 'a') {
          puts("NO");
          return 0;
        }
        nxt[idx] = s[i + 1] - 'a';
      }
    }
  }
  int num = 0;
  string ans;
  for (int i = 0; i < 26; i++) {
    if (can[i] == false) continue;
    if (pre[i] == -1) {
      int x = i;
      while (x != -1) {
        num++;
        ans += (char)x + 'a';
        x = nxt[x];
      }
    }
  }
  if (num != all) {
    puts("NO");
    return 0;
  }
  cout << ans << endl;
  return 0;
}
