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;
using ll = long long;
using vi = vector<int>;
const int maxn = 300006;
int n;
set<int> e[30];
char s[200000];
bool need[30], v[30];
int ans[30], tot;
int ind[30], a[30];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%s", s);
    int len = strlen(s);
    for (int j = 0; j < len; ++j) need[s[j] - 'a'] = true;
    for (int j = 0; j < len - 1; ++j) {
      e[s[j] - 'a'].insert(s[j + 1] - 'a');
      ind[s[j + 1] - 'a']++;
    }
    memset(a, 0, sizeof(a));
    for (int j = 0; j < len; ++j) a[s[j] - 'a']++;
    for (int j = 0; j < 30; ++j)
      if (a[j] > 1) {
        puts("NO");
        return 0;
      }
  }
  memset(v, false, sizeof(v));
  tot = 0;
  for (int i = 0; i < 26; ++i)
    if (ind[i] == 0 && need[i] && !v[i]) {
      int x = i;
      ans[++tot] = i;
      v[x] = true;
      while (e[x].size() != 0) {
        if (e[x].size() > 1) {
          puts("NO");
          return 0;
        }
        x = *e[x].begin();
        ans[++tot] = x;
        if (v[x]) {
          puts("NO");
          return 0;
        }
        v[x] = true;
      }
    }
  for (int i = 0; i < 26; ++i)
    if (need[i] && !v[i]) {
      puts("NO");
      return 0;
    }
  for (int i = 1; i <= tot; ++i) printf("%c", ans[i] + 'a');
  puts("");
}
