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 N = 100031;
vector<int> starts;
int g[500][500], gr[500][500];
string st[N];
int n, cnt_in[N], cnt_out[N];
vector<int> ans;
int need[N];
int ent_here[N];
int used[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> st[i];
    for (int j = 0; j < 26; j++) {
      ent_here[j] = 0;
    }
    for (int j = 0; j < st[i].size(); j++) {
      int id = st[i][j] - 'a';
      ent_here[id]++;
      need[id] = 1;
      if (ent_here[id] > 1) {
        cout << "NO" << endl;
        return 0;
      }
    }
    for (int j = 0; j + 1 < st[i].size(); j++) {
      char a = st[i][j];
      char b = st[i][j + 1];
      a -= 'a';
      b -= 'a';
      g[a][b] = 1;
      gr[b][a] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < 26; j++) {
      if (g[i][j]) {
        cnt_in[j]++;
        cnt_out[i]++;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (cnt_in[i] > 1 || cnt_out[i] > 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (cnt_in[i] == 0 && (cnt_out[i] > 0 || need[i] > 0)) starts.push_back(i);
  }
  for (int i = 0; i < starts.size(); i++) {
    int cur = starts[i];
    while (true) {
      used[cur] = 1;
      ans.push_back(cur);
      if (cnt_out[cur] == 0) break;
      int nxt = -1;
      for (int q = 0; q < 26; q++) {
        if (g[cur][q]) nxt = q;
      }
      cur = nxt;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (need[i] > used[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    cout << char(ans[i] + 'a');
  }
  cout << endl;
  return 0;
}
