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 N = 357;
int n, len;
string s[N];
int poprz[26];
int nast[26];
int wyst[26];
int juz[26];
string res;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  if (n > 351) {
    cout << "NO";
    return 0;
  }
  for (int i = 0; i < 26; ++i) {
    nast[i] = -1;
    poprz[i] = -1;
  }
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
    if (s[i].size() > 26) {
      cout << "NO";
      return 0;
    }
    for (int j = 0; j < s[i].size(); ++j) {
      for (int k = j + 1; k < s[i].size(); ++k) {
        if (s[i][j] == s[i][k]) {
          cout << "NO";
          return 0;
        }
      }
    }
    for (int j = 0; j + 1 < s[i].size(); ++j) {
      if (poprz[s[i][j + 1] - 'a'] != -1 &&
          poprz[s[i][j + 1] - 'a'] != s[i][j] - 'a') {
        cout << "NO";
        return 0;
      }
      poprz[s[i][j + 1] - 'a'] = s[i][j] - 'a';
      if (nast[s[i][j] - 'a'] != -1 &&
          nast[s[i][j] - 'a'] != s[i][j + 1] - 'a') {
        cout << "NO";
        return 0;
      }
      nast[s[i][j] - 'a'] = s[i][j + 1] - 'a';
    }
    for (auto it : s[i]) {
      wyst[it - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (wyst[i]) ++len;
  }
  int curr = 1000000007;
  for (int i = 0; i < len; ++i) {
    if (curr == 1000000007) {
      for (int j = 0; j < 26; ++j) {
        if (wyst[j] && !juz[j] && poprz[j] == -1) curr = min(curr, j);
      }
    }
    if (curr == 1000000007) {
      cout << "NO";
      return 0;
    }
    res += char(curr + 'a');
    juz[curr] = 1;
    if (nast[curr] != -1)
      curr = nast[curr];
    else
      curr = 1000000007;
    if (i == len - 1 && curr != 1000000007) {
      cout << "NO";
      return 0;
    }
  }
  cout << res;
}
