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 MAXN = 1e5 + 10;
char s[MAXN], ans[MAXN];
int nx[30], ind[30], n, len, ansL = 0;
bool vis[30], vis2[30];
int main() {
  for (int i = 0; i < 30; i++) nx[i] = -1;
  scanf("%d", &n);
  bool flag = true;
  for (int j = 0; j < n; j++) {
    scanf("%s", s);
    len = strlen(s);
    for (int i = 1; i < len; i++) {
      int u = s[i - 1] - 'a', v = s[i] - 'a';
      if (nx[u] != -1 && nx[u] != v) flag = 0;
      vis[u] = vis[v] = 1;
      if (nx[u] == -1) nx[u] = v, ind[v]++;
    }
    vis[s[0] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (ind[i] > 1) flag = 0;
  if (!flag) {
    printf("NO");
    exit(0);
  }
  for (int i = 0; i < 26; i++)
    if (vis[i] && !ind[i])
      for (int j = i; j != -1; j = nx[j]) ans[ansL++] = j + 'a', vis2[j] = 1;
  for (int i = 0; i < 26; i++)
    if (vis[i] ^ vis2[i]) flag = 0;
  if (!flag) {
    printf("NO");
    exit(0);
  }
  ans[ansL] = 0;
  printf("%s\n", ans);
  return 0;
}
