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 inf = 1e9 + 1;
const int N = 200123;
int z, n;
char nxt[300];
int in[300];
int vis[300];
set<int> touched;
string ans;
bool solve() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    for (int i = 0; i < s.size() - 1; i++) {
      char c = s[i];
      char d = s[i + 1];
      if (nxt[c] == d) continue;
      if (nxt[c]) {
        return false;
      }
      nxt[c] = d;
      touched.insert(c);
      if (in[d]++) {
        return false;
      }
    }
    touched.insert(s.back());
  }
  queue<int> q;
  for (int c = 'a'; c <= 'z'; c++) {
    if (!in[c] && touched.count(c)) q.push(c);
  }
  while (q.size()) {
    char c = q.front();
    q.pop();
    while (c) {
      if (vis[c]++) return false;
      ans += c;
      c = nxt[c];
    }
  }
  return ans.size() == touched.size();
}
int main() {
  ios_base::sync_with_stdio(0);
  if (!solve()) {
    cout << "NO" << endl;
  } else {
    cout << ans << endl;
  }
}
