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 nex[100], vis[100], ok[100], fib[100], cnt = 0, pre[100];
char s[100010], ans[100];
void print(int x) {
  if (ok[x]) return;
  ok[x] = 1;
  ans[++cnt] = 'a' + x;
  if (nex[x] != -1) print(nex[x]);
}
int main() {
  int n;
  cin >> n;
  memset(nex, -1, sizeof(nex));
  memset(pre, -1, sizeof(pre));
  for (int i = 1; i <= n; i++) {
    scanf("%s", s);
    int len = strlen(s);
    for (int j = 0; j < len - 1; j++) {
      int a = s[j] - 'a', b = s[j + 1] - 'a';
      fib[b] = 1;
      if (nex[a] != -1) {
        if (nex[a] != b) {
          puts("NO");
          return 0;
        }
      } else
        nex[a] = b;
      if (pre[b] != -1) {
        if (pre[b] != a) {
          puts("NO");
          return 0;
        }
      } else
        pre[b] = a;
    }
    for (int j = 0; j < len; j++) vis[s[j] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (!fib[i] && vis[i]) {
      print(i);
    }
  int tot = 0;
  for (int i = 0; i < 26; i++)
    if (vis[i]) tot++;
  if (cnt < tot)
    puts("NO");
  else
    cout << ans + 1 << endl;
  return 0;
}
