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;
int ll = 1;
vector<int> arr(26, -2), used(26, 0);
string ans;
void dfs(int i) {
  char b = (i + 'a');
  ans += b;
  int cnt = 0;
  for (int j = 0; j < 26; j++) {
    if (arr[j] == i) {
      dfs(j);
      ++cnt;
      if (cnt > 1) {
        cout << "NO";
        exit(0);
      }
    }
  }
}
void dfs1(int i) {
  used[i] = 1;
  for (int j = 0; j < 26; j++) {
    if (arr[j] == i) {
      if (used[j] == 1) {
        ll = 0;
        cout << "NO";
        exit(0);
      }
      dfs1(j);
      break;
    }
  }
}
int main() {
  int n;
  cin >> n;
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    set<char> st;
    for (int j = 0; j < s.size(); j++) {
      if (st.count(s[j])) {
        cout << "NO";
        exit(0);
      }
      st.insert(s[j]);
      if (j == 0) {
        if (arr[(s[j] - 'a')] == -2) arr[(s[j] - 'a')] = -1;
        continue;
      }
      if (arr[(s[j] - 'a')] == -2 || arr[(s[j] - 'a')] == -1) {
        arr[(s[j] - 'a')] = (s[j - 1] - 'a');
        continue;
      }
      if (arr[(s[j] - 'a')] != (s[j - 1] - 'a')) {
        cout << "NO";
        exit(0);
      }
    }
  }
  int cnt = 0;
  for (int i = 0; i < 26; i++) {
    if (arr[i] != -2) dfs1(i);
    for (int j = 0; j < 26; j++) used[j] = 0;
  }
  for (int i = 0; i < 26; i++)
    if (arr[i] == -1) {
      dfs(i);
      cnt++;
    }
  if (!cnt) {
    cout << "NO";
    return 0;
  }
  cout << ans;
  return 0;
}
