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 B = 26;
vector<set<int>> g(B);
vector<set<int>> ginv(B);
int color[B];
void DFS_cyc(int v) {
  color[v] = 1;
  if (g[v].size() > 1) {
    cout << "NO";
    exit(0);
  }
  for (auto u : g[v]) {
    if (color[u] == 0) DFS_cyc(u);
    if (color[u] == 1) {
      cout << "NO";
      exit(0);
    }
  }
  color[v] = 2;
}
void cycle() {
  for (int i = 0; i < B; ++i) color[i] = 0;
  for (int i = 0; i < B; ++i)
    if (color[i] == 0) DFS_cyc(i);
}
void DFSinv_cyc(int v) {
  color[v] = 1;
  if (ginv[v].size() > 1) {
    cout << "NO";
    exit(0);
  }
  for (auto u : ginv[v]) {
    if (color[u] == 0) DFSinv_cyc(u);
    if (color[u] == 1) {
      cout << "NO";
      exit(0);
    }
  }
  color[v] = 2;
}
void invcycle() {
  for (int i = 0; i < B; ++i) color[i] = 0;
  for (int i = 0; i < B; ++i)
    if (color[i] == 0) DFSinv_cyc(i);
}
void DFS(int v) {
  color[v] = 1;
  char r = v + 'a';
  cout << r;
  for (auto u : g[v]) {
    if (color[u] == 0) DFS(u);
  }
  color[v] = 2;
}
void solve() {
  int n;
  cin >> n;
  string now;
  int cnt[B], cntt[B];
  for (int i = 0; i < B; ++i) cnt[i] = 0, cntt[i] = 0;
  for (int i = 0; i < n; ++i) {
    cin >> now;
    for (auto p : now) cntt[p - 'a']++;
    for (int j = 0; j < now.size() - 1; ++j)
      g[now[j] - 'a'].insert(now[j + 1] - 'a'), cnt[now[j + 1] - 'a']++;
    for (int j = 1; j < now.size(); ++j)
      ginv[now[j] - 'a'].insert(now[j - 1] - 'a');
  }
  cycle();
  invcycle();
  for (int i = 0; i < B; ++i) color[i] = 0;
  for (int i = 0; i < B; ++i)
    if (color[i] == 0 && cnt[i] == 0 && cntt[i] != 0) DFS(i);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  solve();
  return 0;
}
