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 = 100001, maxM = 26;
char in[maxN];
bool G[maxM][maxM], ll[maxM], vis[maxM];
int f[maxM], g[maxM];
char res[30];
void exit_no() {
  printf("NO\n");
  exit(0);
}
int main() {
  int n;
  scanf("%d", &n);
  while (n--) {
    scanf("%s", in);
    int s = strlen(in);
    for (int i = 1; i < s; i++) G[in[i - 1] - 'a'][in[i] - 'a'] = true;
    if (s == 1) ll[in[0] - 'a'] = true;
  }
  fill(f, f + maxM, -1);
  fill(g, g + maxM, -1);
  for (int i = 0; i < maxM; i++) {
    for (int j = 0; j < maxM; j++) {
      if (G[i][j]) {
        if (f[i] != -1 or g[j] != -1) exit_no();
        f[i] = j, g[j] = i;
      }
    }
  }
  int s = 0;
  for (int i = 0; i < maxM; i++) {
    if (g[i] != -1) continue;
    if (!ll[i] and f[i] == -1) continue;
    for (int v = i; v != -1; v = f[v]) res[s++] = 'a' + v, vis[v] = true;
  }
  res[s] = '\0';
  for (int i = 0; i < maxM; i++)
    if (f[i] != -1 and !vis[i]) exit_no();
  printf("%s\n", res);
  return 0;
}
