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>
#pragma comment(linker, "/STACK:60777216")
using namespace std;
int n;
vector<string> s;
string combine(vector<string> s, int t) {
  if (s.empty()) return "";
  vector<char> res(100, ' ');
  for (int i = (0); i < (s.size()); i++) {
    int pos = 0;
    for (int j = (0); j < (s[i].size()); j++)
      if (s[i][j] - 'a' == t) pos = j;
    for (int j = (0); j < (s[i].size()); j++) {
      if (res[50 + j - pos] != ' ' && res[50 + j - pos] != s[i][j]) return "NO";
      res[50 + j - pos] = s[i][j];
    }
  }
  string w;
  for (int i = (0); i < (100); i++)
    if (res[i] != ' ') w.push_back(res[i]);
  int mask = 0;
  for (int i = (0); i < (w.size()); i++) {
    if (mask & (1 << (w[i] - 'a'))) return "NO";
    mask |= 1 << (w[i] - 'a');
  }
  return w;
}
int main() {
  cin >> n;
  string t;
  getline(cin, t);
  for (int i = (0); i < (n); i++) {
    getline(cin, t);
    int mask = 0;
    for (int j = (0); j < (t.size()); j++) {
      if (mask & (1 << (t[j] - 'a'))) {
        puts("NO");
        return 0;
      }
      mask |= 1 << (t[j] - 'a');
    }
    s.push_back(t);
  }
  for (int i = (0); i < (26); i++) {
    vector<string> ns, tmp;
    for (int j = (0); j < (s.size()); j++) {
      bool has = 0;
      for (int k = (0); k < (s[j].size()); k++)
        if (s[j][k] - 'a' == i) has = 1;
      if (has)
        tmp.push_back(s[j]);
      else
        ns.push_back(s[j]);
    }
    string t = combine(tmp, i);
    if (t == "NO") {
      puts("NO");
      return 0;
    }
    if (!t.empty()) ns.push_back(t);
    s = ns;
  }
  sort((s).begin(), (s).end());
  for (int i = (0); i < (s.size()); i++) cout << s[i];
  puts("");
  return 0;
}
