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 = 26;
const long long M = 1e9 + 7;
long long n, k;
vector<int> f;
vector<string> s;
vector<int> nxt(N, -1);
vector<int> prv(N, -1);
bool zx[N];
int main() {
  cin >> n;
  s.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    int l = s[i].length();
    vector<bool> qw(N, 0);
    for (int j = 0; j < l; j++) {
      int p = s[i][j] - 'a';
      if (qw[p]) {
        return cout << "NO", 0;
      }
      qw[p] = 1;
      zx[p] = 1;
      if (j != l - 1) {
        if (nxt[p] != s[i][j + 1] - 'a' && nxt[p] != -1) return cout << "NO", 0;
        nxt[p] = s[i][j + 1] - 'a';
      }
      if (j != 0) {
        if (prv[p] != s[i][j - 1] - 'a' && prv[p] != -1) return cout << "NO", 0;
        prv[p] = s[i][j - 1] - 'a';
      }
    }
  }
  string rez = "";
  for (int i = 0; i < N; i++) {
    if (prv[i] == -1 && zx[i]) {
      int y = i;
      while (y != -1) {
        rez += char(y + 'a');
        y = nxt[y];
      }
    }
  }
  int g = 0;
  for (int i = 0; i < N; i++)
    if (zx[i]) g++;
  if (rez.length() != g) return cout << "NO", 0;
  cout << rez;
  return 0;
}
