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;
bool jest[30];
int graf[30];
bool vis[30];
bool dfsCykl(int a, int x) {
  if (graf[a] == a) return false;
  if (graf[a] == x) return true;
  return dfsCykl(graf[a], x);
}
string dfs(int a) {
  vis[a] = true;
  if (graf[a] == a) {
    char tmp = 'a' + a;
    string res;
    res += tmp;
    return res;
  }
  string res = dfs(graf[a]);
  return (char)(a + 'a') + res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < 26; i++) graf[i] = i;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    vector<bool> bylo(26);
    for (auto c : s) {
      if (bylo[c - 'a']) {
        cout << "NO\n";
        return 0;
      }
      bylo[c - 'a'] = true;
      jest[c - 'a'] = true;
    }
    for (int j = 0; j < (int)(s).size() - 1; j++) {
      if (graf[s[j] - 'a'] == s[j] - 'a') {
        graf[s[j] - 'a'] = s[j + 1] - 'a';
      } else if (graf[s[j] - 'a'] != s[j + 1] - 'a') {
        cout << "NO\n";
        return 0;
      }
    }
  }
  vector<int> deg(30);
  for (int i = 0; i < 26; i++)
    if (graf[i] != i) {
      if (deg[graf[i]]) {
        cout << "NO\n";
        return 0;
      } else
        deg[graf[i]]++;
    }
  for (int i = 0; i < 26; i++) {
    if (graf[i] != i && dfsCykl(i, i)) {
      cout << "NO\n";
      return 0;
    }
  }
  vector<string> odp;
  for (int i = 0; i < 26; i++) {
    if (graf[i] != i && deg[i] == 0) {
      odp.push_back(dfs(i));
    }
  }
  for (int i = 0; i < 26; i++) {
    if (jest[i] && !vis[i]) {
      char c = i + 'a';
      string tmp;
      tmp += c;
      odp.push_back(tmp);
    }
  }
  sort((odp).begin(), (odp).end());
  for (auto s : odp) cout << s;
  cout << "\n";
  return 0;
}
