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;
struct point {
  bool is;
  char val;
  char prev;
  char next;
};
int main() {
  int i, n, cnt;
  scanf("%d ", &n);
  map<char, point> v;
  for (i = 0; i < 26; i++) {
    v[i + 'a'].prev = 0;
    v[i + 'a'].next = 0;
  }
  string s;
  set<char> a;
  for (i = 0; i < n; i++) {
    cin >> s;
    a.clear();
    for (int j = 0; j < s.size(); j++) {
      if (a.count(s[j])) {
        printf("NO");
        return 0;
      } else
        a.insert(s[j]);
      if (v[s[j]].is) {
        if (j > 0) {
          if (v[s[j]].prev == 0) v[s[j]].prev = s[j - 1];
          if (v[s[j]].prev != s[j - 1]) {
            printf("NO");
            return 0;
          }
        }
        if (j < s.size() - 1) {
          if (v[s[j]].next == 0) v[s[j]].next = s[j + 1];
          if (v[s[j]].next != s[j + 1]) {
            printf("NO");
            return 0;
          }
        }
      }
      if (!v[s[j]].is) {
        v[s[j]].val = s[j];
        if (j > 0) {
          if (v[s[j]].prev == 0) v[s[j]].prev = s[j - 1];
          if (v[s[j]].prev != s[j - 1]) {
            printf("NO");
            return 0;
          }
        }
        if (j < s.size() - 1) {
          if (v[s[j]].next == 0) v[s[j]].next = s[j + 1];
          if (v[s[j]].next != s[j + 1]) {
            printf("NO");
            return 0;
          }
        }
        v[s[j]].is = 1;
      }
    }
  }
  vector<char> ans;
  cnt = 0;
  int cnt2 = 0;
  for (i = 0; i < 26; i++)
    if (v['a' + i].is) {
      if (v['a' + i].prev == 0) ans.push_back('a' + i);
      cnt2++;
    }
  if (ans.size() == 0) {
    printf("NO");
    return 0;
  }
  sort(ans.begin(), ans.end());
  for (i = 0; i < ans.size(); i++) {
    point p;
    cnt++;
    for (p = v[ans[i]]; p.next != 0; p = v[p.next]) cnt++;
  }
  if (cnt != cnt2) {
    printf("NO");
    return 0;
  }
  for (i = 0; i < ans.size(); i++) {
    point p;
    for (p = v[ans[i]]; p.next != 0; p = v[p.next]) printf("%c", p.val);
    printf("%c", p.val);
  }
  return 0;
}
