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 int MAX = 1e5 + 5;
string s[MAX];
set<int> in[MAX];
set<int> out[MAX];
int vis[MAX];
string st;
void dfs(int s) {
  st += (s + 'a');
  if (!out[s].empty()) {
    dfs(*out[s].begin());
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < s[i].length() - 1; j++) {
      out[s[i][j] - 'a'].insert(s[i][j + 1] - 'a');
      in[s[i][j + 1] - 'a'].insert(s[i][j] - 'a');
    }
    for (int j = 0; j < s[i].length(); j++) {
      vis[s[i][j] - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (out[i].size() > 1 || in[i].size() > 1) {
      printf("NO\n");
      return 0;
    }
  }
  int cnt = 0;
  for (int i = 0; i < 26; i++) {
    if (vis[i]) {
      cnt++;
    }
  }
  string ans = "";
  for (int i = 0; i < 26; i++) {
    if (vis[i] && in[i].size() == 0) {
      st = "";
      dfs(i);
      ans += st;
    }
  }
  if (ans.length() < cnt) {
    printf("NO\n");
  } else {
    cout << ans;
  }
  return 0;
}
