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;
bool vis[26];
string ans;
int e1[26], ind[26];
bitset<26> e[26];
void DFS(int u) {
  if (u == -1) {
    return;
  }
  vis[u] = 0;
  ans += u + 'a';
  DFS(e1[u]);
}
int main() {
  int i, j, sz, a, b;
  string s;
  ios_base::sync_with_stdio(0);
  memset(e1, -1, sizeof(e1));
  cin >> n;
  for (i = (1); i <= (n); i++) {
    cin >> s;
    sz = s.length();
    for (j = (0); j <= (sz - 2); j++) {
      a = s[j] - 'a';
      b = s[j + 1] - 'a';
      vis[a] = vis[b] = 1;
      if (!e[a][b]) {
        e[a][b] = 1;
        ind[b]++;
        e1[a] = b;
      }
    }
    vis[s[0] - 'a'] = 1;
  }
  for (i = (0); i <= (25); i++) {
    if (e[i].count() > 1 || ind[i] > 1) {
      cout << "NO";
      exit(0);
    }
  }
  ans = "";
  for (i = (0); i <= (25); i++) {
    if (!ind[i] && vis[i]) {
      DFS(i);
    }
  }
  for (i = (0); i <= (25); i++) {
    if (vis[i]) {
      cout << "NO";
      exit(0);
    }
  }
  cout << ans;
  return 0;
}
