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;
void GO() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
const long long INF = 1e9 + 7;
const long long mod = (1e9) + 7;
const int maxn = (int)2e5 + 1;
const long double PI = acos(-1.0);
const long double eps = 0.000000001;
mt19937 rnd(time(0));
int g[27], u[27], us[27], p[27];
vector<int> top_sort;
void dfs(int v) {
  if (v == -1 || u[v] == 2) return;
  if (u[v] == 1) {
    cout << "NO\n";
    exit(0);
  }
  u[v] = 1;
  dfs(g[v]), top_sort.push_back(v);
  u[v] = 2;
}
void solve() {
  int n;
  cin >> n;
  memset(g, -1, sizeof(g));
  memset(p, -1, sizeof(p));
  memset(u, 0, sizeof(u));
  memset(us, 0, sizeof(us));
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (int j = 0; j < int(s.size()) - 1; ++j) {
      int v = s[j] - 'a';
      int to = s[j + 1] - 'a';
      us[v] = 1;
      if (g[v] == -1) {
        g[v] = to;
        if (p[to] == -1) p[to] = v;
      }
      if (g[v] != to || p[to] != v) {
        cout << "NO\n";
        return;
      }
    }
    us[int(s.back() - 'a')] = 1;
  }
  vector<string> ans;
  for (int i = 0; i < 27; ++i)
    if (us[i]) dfs(i);
  memset(u, 0, sizeof(u));
  for (int i = int(top_sort.size()) - 1; i >= 0; --i) {
    int v = top_sort[i];
    if (u[v]) continue;
    string plus;
    while (v != -1) {
      plus += char(v + 'a');
      u[v] = 1, v = g[v];
    }
    ans.push_back(plus);
  }
  sort(ans.begin(), ans.end());
  for (auto &i : ans) cout << i;
}
signed main() {
  GO();
  int Q = 1;
  while (Q--) {
    solve();
  }
}
