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 dict = 26;
int next2[dict], prev2[dict];
bool used2[dict];
bool usedInDfs2[dict];
string dfs(int ch) {
  usedInDfs2[ch] = true;
  string ret;
  ret.push_back('a' + ch);
  if (next2[ch] != -1) {
    if (usedInDfs2[next2[ch]])
      return ret += "-";
    else
      ret += dfs(next2[ch]);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  for (int i = 0; i < 26; i++) {
    next2[i] = -1;
    prev2[i] = -1;
    used2[i] = false;
    usedInDfs2[i] = false;
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < s.size(); j++) {
      int prevCode = -1;
      if (j > 0) prevCode = s[j - 1] - 'a';
      int code = s[j] - 'a';
      int nextCode = -1;
      if (j + 1 < s.size()) nextCode = s[j + 1] - 'a';
      used2[code] = true;
      if (j > 0) {
        if (prev2[code] == -1) {
          prev2[code] = prevCode;
        } else if (prevCode == -1) {
        } else if (prevCode != prev2[code]) {
          cout << "NO";
          return 0;
        }
      }
      if (j + 1 < s.size()) {
        if (next2[code] == -1) {
          next2[code] = nextCode;
        } else if (nextCode == -1) {
        } else if (nextCode != next2[code]) {
          cout << "NO";
          return 0;
        }
      }
    }
  }
  string ans;
  for (int i = 0; i < dict; i++) {
    if (prev2[i] == -1 && used2[i] && !usedInDfs2[i]) {
      string add = dfs(i);
      bool er = false;
      for (char c : add)
        if (c == '-') er = true;
      if (er) {
        cout << "NO";
        return 0;
      }
      ans += add;
    }
  }
  for (int i = 0; i < dict; i++) {
    if (used2[i] && !usedInDfs2[i]) {
      cout << "NO";
      return 0;
    }
  }
  cout << ans;
  return 0;
}
