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 = 30 + 3;
int n, deg[N];
bool exist[N], match[N][N], vis[N], foo = true;
vector<int> adj[N];
vector<string> V;
void init() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void dfs(int u, string s) {
  vis[u] = 1;
  s += (char)(u + 'a');
  cerr << u << ' ' << s << ' ' << adj[u].size() << '\n';
  if (adj[u].size() == 0) {
    V.push_back(s);
    return;
  }
  for (int v : adj[u])
    if (!vis[v])
      dfs(v, s);
    else
      foo = false;
}
void solve() {
  cin >> n;
  for (int i = 1; i <= (int)n; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j <= (int)s.size() - 1; ++j) {
      exist[s[j] - 'a'] = 1;
      if (j < s.size() - 1) match[s[j] - 'a'][s[j + 1] - 'a'] = 1;
    }
  }
  for (int i = 0; i <= (int)25; ++i)
    for (int j = 0; j <= (int)25; ++j)
      if (match[i][j]) adj[i].push_back(j), deg[j]++;
  for (int i = 0; i <= (int)25; ++i)
    if (adj[i].size() > 1) foo = false;
  for (int i = 0; i <= (int)25; ++i)
    for (int v : adj[i])
      for (int k : adj[v])
        if (i == k) foo = false;
  for (int i = 0; i <= (int)25; ++i)
    if (deg[i] == 0 && exist[i]) dfs(i, "");
  for (int i = 0; i <= (int)25; ++i)
    if (exist[i] && !vis[i]) foo = false;
  if (!foo)
    cout << "NO";
  else
    for (auto u : V) cout << u;
}
int main() {
  init();
  solve();
}
