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;
constexpr int C = 128;
bool seen[C];
char pre[C], nxt[C];
int main() {
  int n;
  scanf("%d", &n);
  getchar();
  for (auto i = (0); i < (n); i++) {
    char a = 0, b;
    while ((b = getchar()) != '\n') {
      if (a != 0) {
        char x = b;
        while (x && x != a) x = nxt[x];
        if (x || (nxt[a] && nxt[a] != b) || (pre[b] && pre[b] != a)) {
          printf("NO\n");
          return 0;
        }
        nxt[a] = b, pre[b] = a;
      }
      seen[b] = true, a = b;
    }
  }
  for (auto c = ('a'); c < ('z' + 1); c++)
    if (seen[c] && !pre[c])
      for (char x = c; x; x = nxt[x]) putchar(x);
  putchar('\n');
  return 0;
}
