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 mod = 1e9 + 7;
const int maxn = 1e5 + 5;
string s[maxn];
bool vis[30], fv[30];
bool g[30][30];
int d[30], f[30];
string dfs(int u) {
  string s = "";
  s += (char)('a' + u);
  for (int v = 0; v < 26; v++) {
    if (g[u][v]) s += dfs(v);
  }
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, i, j;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> s[i];
    memset(vis, 0, sizeof(vis));
    m = s[i].size();
    for (j = 0; j < m; j++) {
      fv[s[i][j] - 'a'] = 1;
      if (vis[s[i][j] - 'a'])
        return puts("NO"), 0;
      else
        vis[s[i][j] - 'a'] = 1;
    }
    for (j = 1; j < m; j++) g[s[i][j - 1] - 'a'][s[i][j] - 'a'] = 1;
  }
  for (i = 0; i < 26; i++) {
    for (j = 0; j < 26; j++) {
      d[j] += g[i][j];
      f[i] += g[i][j];
    }
  }
  for (i = 0; i < 26; i++) {
    if (f[i] > 1 || d[i] > 1) return puts("NO"), 0;
  }
  string ans = "";
  for (i = 0, j = 0; i < 26; i++) {
    j += fv[i];
    if (fv[i] && d[i] == 0) ans += dfs(i);
  }
  if (ans.size() != j)
    puts("NO");
  else
    cout << ans << endl;
  return 0;
}
