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;
int n, nxt[30], vis[30], par[30], is[30];
vector<string> vs;
char ans[300007], buf[100003];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", buf);
    vs.push_back(buf);
    int m = (int)strlen(buf);
    sort(buf, buf + m);
    if ((int)(unique(buf, buf + m) - buf) < m) {
      puts("NO");
      return 0;
    }
    for (int j = 0; j < m; j++) {
      is[buf[j] - 'a'] = 1;
    }
  }
  memset(nxt, -1, sizeof(nxt));
  for (auto &s : vs) {
    for (int i = 0; i < (int)s.size() - 1; i++) {
      if (nxt[s[i] - 'a'] >= 0 && nxt[s[i] - 'a'] != s[i + 1] - 'a') {
        puts("NO");
        return 0;
      }
      nxt[s[i] - 'a'] = s[i + 1] - 'a';
      par[s[i + 1] - 'a'] = 1;
    }
  }
  vector<string> ans;
  vector<int> sts;
  for (int i = 0; i < 26; i++) {
    if (is[i] && !par[i]) sts.push_back(i);
  }
  for (auto &st : sts) {
    int cur = st;
    string tmp;
    do {
      tmp.push_back(cur + 'a');
      vis[cur] = 1;
      cur = nxt[cur];
      if (vis[cur]) {
        puts("NO");
        return 0;
      }
    } while (cur != -1);
    ans.push_back(tmp);
  }
  for (int i = 0; i < 26; i++) {
    if (is[i] && !vis[i]) {
      puts("NO");
      return 0;
    }
  }
  sort(ans.begin(), ans.end());
  for (auto &s : ans) {
    ;
    printf("%s", s.c_str());
  }
  puts("");
  return 0;
}
