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 long double pi = acosl(-1);
const int maxn = 200010;
const long long mod = 1e6 + 3;
const int inf = 1e9;
int n;
char s[maxn];
int a[26][26];
vector<int> g[26];
int vis[26], vis2[26], deg[26];
set<string> vec;
string str;
void die() {
  printf("NO\n");
  exit(0);
}
void dfs(int i) {
  if (vis2[i] || g[i].size() > 1) die();
  vis2[i] = 1;
  str += char(i + 'a');
  if (!g[i].empty()) dfs(g[i][0]);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", s);
    int len = strlen(s);
    for (int j = 0; j < len - 1; j++)
      a[s[j] - 'a'][s[j + 1] - 'a'] = 1, vis[s[j] - 'a'] = 1;
    vis[s[len - 1] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    for (int j = 0; j < 26; j++) {
      if (a[i][j]) g[i].push_back(j), deg[j]++;
    }
  for (int i = 0; i < 26; i++) {
    if (!deg[i] && !g[i].empty()) {
      str = "";
      dfs(i);
      vec.insert(str);
    }
  }
  for (int i = 0; i < 26; i++) {
    if (vis[i] != vis2[i]) {
      if (deg[i])
        die();
      else
        vec.insert(string(1, char(i + 'a')));
    }
  }
  for (auto s : vec) cout << s;
  return 0;
}
