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;
class Solution {
  int n;
  int nxt[26], pev[26];
  bool inst[26] = {0}, vis[26] = {0}, occ[26] = {0};
  string ans, s;
  void dfs(int v) {
    if (inst[v]) {
      cout << "NO\n";
      exit(0);
    }
    ans.push_back(v + 'a');
    vis[v] = true;
    inst[v] = true;
    if (nxt[v] != -1) dfs(nxt[v]);
    inst[v] = false;
  }

 public:
  void run() {
    memset(nxt, -1, sizeof(nxt));
    memset(pev, -1, sizeof(pev));
    cin >> n;
    while (n--) {
      cin >> s;
      for (int i = 0; i < s.size() - 1; ++i) {
        if (nxt[s[i] - 'a'] != -1 and nxt[s[i] - 'a'] != s[i + 1] - 'a')
          return cout << "NO\n", void();
        nxt[s[i] - 'a'] = s[i + 1] - 'a';
        if (pev[s[i + 1] - 'a'] != -1 and pev[s[i + 1] - 'a'] != s[i] - 'a')
          return cout << "NO\n", void();
        pev[s[i + 1] - 'a'] = s[i] - 'a';
      }
      for (char c : s) occ[c - 'a'] = 1;
    }
    for (int i = 0; i < 26; ++i)
      if (pev[i] == -1 and not vis[i] and occ[i]) dfs(i);
    if (ans.size() != count(occ, occ + 26, true)) return cout << "NO\n", void();
    cout << ans << '\n';
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  Solution().run();
  return 0;
}
