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;
bool vis[26], can[26];
int nt[26], pv[26];
char s[100005];
int main() {
  memset(nt, -1, sizeof(nt));
  memset(pv, -1, sizeof(pv));
  int n;
  scanf("%d", &n);
  bool flag = 1;
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    int m = strlen(s);
    for (int j = 0; j < m; j++) can[s[j] - 'a'] = 1;
    for (int j = 1; j < m; j++) {
      int x = s[j - 1] - 'a', y = s[j] - 'a';
      if (nt[x] != -1 && nt[x] != y) flag = 0;
      if (pv[y] != -1 && pv[y] != x) flag = 0;
      nt[x] = y;
      pv[y] = x;
    }
  }
  if (!flag)
    printf("NO\n");
  else {
    vector<char> ans;
    for (int i = 0; i < 26; i++) {
      if (can[i] && pv[i] == -1) {
        int u = i;
        vis[u] = 1;
        ans.emplace_back(u + 'a');
        while (nt[u] != -1) {
          if (vis[nt[u]]) {
            printf("NO\n");
            return 0;
          }
          u = nt[u];
          vis[u] = 1;
          ans.emplace_back(u + 'a');
        }
      }
    }
    for (int i = 0; i < 26; i++) {
      if (can[i] && !vis[i]) {
        printf("NO\n");
        return 0;
      }
    }
    for (auto it : ans) printf("%c", it);
    printf("\n");
  }
}
