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 N = 200200;
const int mod = 1e9 + 7;
int n;
string s[N];
int cnt[26];
int rec[26];
int nxt[26];
char t[100100];
int used[26];
int deg[26];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", t);
    s[i] = t;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j < s[i].length(); ++j) {
      if (cnt[s[i][j] - 'a'] == i) return puts("NO"), 0;
      cnt[s[i][j] - 'a'] = i;
    }
  }
  for (int i = 0; i < 26; ++i) nxt[i] = -1, rec[i] = -1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j + 1 < s[i].length(); ++j) {
      if (nxt[s[i][j] - 'a'] != -1 && nxt[s[i][j] - 'a'] != s[i][j + 1] - 'a')
        return puts("NO"), 0;
      nxt[s[i][j] - 'a'] = s[i][j + 1] - 'a';
      if (rec[s[i][j + 1] - 'a'] != -1 &&
          rec[s[i][j + 1] - 'a'] != s[i][j] - 'a')
        return puts("NO"), 0;
      rec[s[i][j + 1] - 'a'] = s[i][j] - 'a';
      ++deg[s[i][j + 1] - 'a'];
    }
  string res;
  while (true) {
    int id = -1, c = 0;
    for (int i = 0; i < 26; ++i)
      if (cnt[i] && !used[i]) {
        c++;
        if (deg[i] == 0 && id == -1) id = i;
      }
    if (c == 0) break;
    if (id == -1) return puts("NO"), 0;
    while (id != -1 && !used[id])
      used[id] = 1, res += char(id + 'a'), id = nxt[id];
  }
  puts(res.c_str());
  return 0;
}
