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 long ONE = 1;
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
const int mn = 2e5 + 34;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
int nxt[mn];
bool vis[mn], st[mn];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n;
  memset(vis, 1, sizeof(vis));
  memset(st, 1, sizeof(st));
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (char c : s) vis[c] = 0;
    for (int i = 0; i + 1 < s.size(); i++) {
      if (nxt[s[i]] && nxt[s[i]] != s[i + 1]) {
        printf("NO\n");
        return 0;
      }
      nxt[s[i]] = s[i + 1];
      st[s[i + 1]] = 0;
    }
  }
  string ans;
  for (int i = 'a'; i <= 'z'; i++)
    if (!vis[i] && st[i]) {
      for (int j = i; j; j = nxt[j]) {
        if (vis[j]) {
          printf("NO\n");
          return 0;
        }
        vis[j] = 1;
        ans.push_back(j);
      }
    }
  for (int i = 'a'; i <= 'z'; i++)
    if (!vis[i]) {
      printf("NO\n");
      return 0;
    }
  printf("%s", ans.c_str());
}
