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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
const long double Eps = 1e-9;
int dx[9] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
int dy[9] = {0, 0, 0, 1, -1, 1, -1, 1, -1};
using namespace std;
vector<int> g[26];
void TestCase() {
  int n;
  cin >> n;
  vector<string> v(n);
  vector<int> f(26), take(26, 0);
  bool ok = true;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    f.assign(26, 0);
    for (int j = 0; j < (int)v[i].size(); j++) f[v[i][j] - 'a']++;
    for (int j = 0; j < 26; j++) ok &= (f[j] <= 1);
  }
  if (ok) {
    string t;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < (int)v[i].size(); j++) g[v[i][j] - 'a'].push_back(i);
    }
    map<int, int> p, q;
    int cnt = 0;
    while (1) {
      if (p.empty()) {
        char st = '-';
        for (int i = 0; i < 26; i++) {
          if ((!take[i]) && (!g[i].empty())) {
            bool yes = true;
            for (auto u : g[i]) {
              if (v[u][0] != char(i + 'a')) yes = false;
            }
            if (yes) {
              st = char(i + 'a');
              break;
            }
          }
        }
        if (st == '-') {
          for (int i = 0; i < 26; i++) {
            if ((!take[i]) && (!g[i].empty())) {
              ok = false;
              break;
            }
          }
          break;
        } else {
          t.push_back(st);
          take[st - 'a'] = 1;
          for (auto u : g[st - 'a']) p.insert({u, 0});
        }
      }
      for (auto &u : p) {
        if (u.second + 1 < (int)v[u.first].size())
          q.insert({u.first, u.second + 1});
      }
      if (!q.empty()) {
        int sid = q.begin()->first, cid = q.begin()->second;
        char ch = v[sid][cid];
        for (auto &o : q) {
          if (ch != v[o.first][o.second]) ok = false;
        }
        if (!ok) break;
        for (auto u : g[ch - 'a']) {
          if (v[u][0] == ch)
            q.insert({u, 0});
          else {
            if (q.find(u) == q.end()) ok = false;
          }
        }
        if (!ok)
          break;
        else {
          t.push_back(ch);
          take[ch - 'a'] = 1;
        }
      }
      p = q;
      q.clear();
    }
    ok ? cout << t : cout << "NO";
  } else
    cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int T = 1;
  while (T--) {
    TestCase();
    cout << "\n";
  }
  return 0;
}
