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;
int n, i, j, cnt;
vector<int> d(100);
bool flag = 0;
vector<vector<char> > w(30, vector<char>(30));
vector<char> h(100), used(100000);
vector<int> g[100];
vector<pair<int, int> > p;
string s[100000], answer[100];
int dfs(int v);
int dfs2(int v);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> s[i];
    d.assign(30, false);
    for (j = 0; j < s[i].size(); j++) {
      d[s[i][j] - 'a']++;
      if (j) {
        if (w[s[i][j - 1] - 'a'][s[i][j] - 'a'] == false) {
          w[s[i][j - 1] - 'a'][s[i][j] - 'a'] = true;
          g[s[i][j - 1] - 'a'].push_back(s[i][j] - 'a');
        }
      }
    }
    for (j = 0; j < 26; j++) {
      if (d[j] > 1) {
        cout << "NO";
        return 0;
      }
      if (d[j]) h[j] = 1;
    }
  }
  for (i = 0; i < 26; i++) {
    used.assign(30, false);
    cnt = 0;
    if (h[i]) {
      dfs(i);
      p.push_back(make_pair(cnt, i));
    }
  }
  used.assign(30, false);
  sort(p.begin(), p.end(), greater<pair<int, int> >());
  j = 0;
  for (i = 0; i < p.size(); i++) {
    if (!used[p[i].second]) {
      if (!dfs2(p[i].second)) {
        cout << "NO";
        return 0;
      }
      j++;
    }
  }
  if (flag) {
    cout << "NO";
    return 0;
  }
  sort(answer, answer + j);
  for (i = 0; i < j; i++) cout << answer[i];
  return 0;
}
int dfs(int v) {
  used[v] = true;
  cnt++;
  for (int i = 0; i < g[v].size(); i++) {
    if (!used[g[v][i]]) dfs(g[v][i]);
  }
}
int dfs2(int v) {
  used[v] = true;
  answer[j] += char(v + 'a');
  if (g[v].size() > 1) flag = true;
  for (int i = 0; i < g[v].size(); i++) {
    if (!used[g[v][i]])
      dfs2(g[v][i]);
    else
      flag = true;
  }
  return 1;
}
