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:167772160000")
using namespace std;
char curLetters[26];
void repLet(const string &a) {
  for (int j = 0; j < (26); j++) curLetters[j] = false;
  for (auto x : a) {
    if (curLetters[x - 'a']) {
      cout << "NO" << endl;
      exit(0);
    }
    curLetters[x - 'a'] = true;
  }
}
string s[110000];
string neededStrings[30];
string un(string a, string b) {
  int positions[30];
  for (int i = 0; i < (26); i++) positions[i] = -1;
  for (int i = 0; i < (a.length()); i++) {
    positions[a[i] - 'a'] = i;
  }
  pair<int, int> con;
  for (int i = 0; i < (b.length()); i++) {
    if (positions[b[i] - 'a'] == -1) continue;
    con = pair<int, int>(positions[b[i] - 'a'], i);
  }
  int i, j;
  int begLen = max(con.first, con.second);
  i = con.first - begLen;
  j = con.second - begLen;
  string ret = "";
  for (; i < a.length() || j < b.length(); i++, j++) {
    if (i < 0 || i >= a.length())
      ret += b[j];
    else if (j < 0 || j >= b.length()) {
      ret += a[i];
    } else {
      if (a[i] != b[j]) {
        cout << "NO";
        exit(0);
      }
      ret += a[i];
    }
  }
  repLet(ret);
  return ret;
}
bool haveSame(const string &a, const string &b) {
  int positions[30];
  for (int i = 0; i < (26); i++) positions[i] = -1;
  for (int i = 0; i < (a.length()); i++) {
    positions[a[i] - 'a'] = i;
  }
  for (int i = 0; i < (b.length()); i++)
    if (positions[b[i] - 'a'] != -1) return true;
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < (n); i++) {
    cin >> s[i];
    repLet(s[i]);
  }
  vector<string> curStrings;
  for (int i = 0; i < (n); i++) {
    string cur = s[i];
    for (int j = 0; j < (curStrings.size()); j++) {
      if (haveSame(cur, curStrings[j])) {
        cur = un(cur, curStrings[j]);
        curStrings[j] = curStrings.back();
        curStrings.pop_back();
        j--;
      }
    }
    curStrings.push_back(cur);
  }
  sort(curStrings.begin(), curStrings.end());
  for (auto x : curStrings) cout << x;
  return 0;
}
