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 target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int N = 26;
vector<int> g[N];
int color[N], pr[N], u[N], have[N];
set<int> degin[N], degout[N];
string cur;
vector<string> ans;
bool dfs1(int v) {
  color[v] = true;
  for (auto& x : g[v])
    if (color[x] == 0) {
      pr[x] = v;
      if (dfs1(x)) return true;
    } else if (color[x] == 1) {
      cout << "NO\n";
      exit(0);
    }
  color[v] = 2;
  return false;
}
void dfs2(int v) {
  u[v] = true;
  cur += char(v + 'a');
  for (auto& x : g[v])
    if (!u[x]) dfs2(x);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n = 0;
  cin >> n;
  string s;
  for (int i = 0; i < n; i++) {
    cin >> s;
    have[s[0] - 'a']++;
    for (int j = 1; j < s.size(); j++) {
      have[s[j] - 'a']++;
      g[s[j - 1] - 'a'].push_back(s[j] - 'a');
      degin[s[j] - 'a'].insert(s[j - 1]);
    }
  }
  for (int i = 0; i < 26; i++) {
    for (auto& x : g[i]) degout[i].insert(x);
    if (degin[i].size() > 1 || degout[i].size() > 1) {
      cout << "NO\n";
      return 0;
    }
  }
  for (int i = 0; i < 26; i++)
    if (color[i] == 0 && dfs1(i)) break;
  for (int i = 0; i < 26; i++)
    if (degin[i].size() == 0 && (!g[i].empty() || have[i])) {
      cur.clear();
      dfs2(i);
      ans.push_back(cur);
    }
  sort((ans).begin(), (ans).end());
  for (auto& x : ans) cout << x;
  return 0;
}
