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;
string str[100010], ans;
int cnt[30], start, mat[30][30], in[30], out[30], ase[30], vis[30];
int valid(string str) {
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i <= (int)str.size() - 1; i++) {
    int a = str[i] - 'a';
    if (cnt[a]) return 1;
    cnt[a]++;
  }
  return 0;
}
void up(string str) {
  int n = str.size(), a, b;
  for (int i = 0; i <= n - 1; i++) {
    a = str[i] - 'a';
    ase[a] = 1;
    if (i < n - 1) {
      b = str[i + 1] - 'a';
      if (mat[a][b] == 0) mat[a][b]++, out[a]++, in[b]++;
    }
  }
}
int ck;
string ans3;
void dfs(int p) {
  char ch = p + 'a';
  ans += ch;
  ck = 0;
  vis[p] = 1;
  for (int i = 0; i <= 25; i++)
    if (mat[p][i] == 1) dfs(i);
}
int main() {
  ios::sync_with_stdio(false);
  int i, j, n, m, x, p, a, bm, c = 0;
  cin >> n;
  for (int i = 0; i <= n - 1; i++)
    cin >> str[i], up(str[i]), c = max(x, valid(str[i]));
  for (int i = 0; i <= 25; i++) {
    if (in[i] > 1 || out[i] > 1 || c == 1) {
      cout << "NO";
      return 0;
    }
  }
  ck = 1;
  for (int i = 0; i <= 25; i++)
    if (ase[i] == 1 && in[i] == 0) dfs(i);
  for (int i = 0; i <= 25; i++)
    if (vis[i] != ase[i]) ck = 1;
  if (ck)
    cout << "NO";
  else
    cout << ans << '\n';
  return 0;
}
