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;
char str[100010];
bool line[27][27];
bool p[27], vis[27];
int in[27], out[27];
string ans;
bool dfs(int value);
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", str);
    int len = strlen(str);
    for (int i = 0; i < len - 1; i++) {
      if (str[i] == str[i + 1]) {
        printf("NO\n");
        return 0;
      }
      line[str[i] - 'a'][str[i + 1] - 'a'] = true;
    }
    if (len == 1) p[str[0] - 'a'] = true;
  }
  ans = "";
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < 26; j++)
      if (line[i][j]) {
        out[i]++;
        in[j]++;
      }
  }
  for (int i = 0; i < 26; i++)
    if (in[i] > 1 || out[i] > 1) {
      printf("NO\n");
      return 0;
    }
  for (int i = 0; i < 26; i++) {
    if (out[i] != 0 && in[i] == 0)
      if (!dfs(i)) {
        printf("NO\n");
        return 0;
      }
    if (p[i] && in[i] == 0 && out[i] == 0) ans += (i + 'a');
  }
  for (int i = 0; i < 26; i++)
    if (in[i] != 0 && out[i] != 0 && !vis[i]) {
      printf("NO\n");
      return 0;
    }
  cout << ans << endl;
  return 0;
}
bool dfs(int value) {
  vis[value] = true;
  ans += (value + 'a');
  for (int i = 0; i < 26; i++) {
    if (line[value][i]) {
      return dfs(i);
    }
  }
  return true;
}
