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, l, r;
int pos[26];
char ans[65];
string s[100005];
queue<int> Q;
vector<int> vol[26];
vector<string> final_ans;
bool vis[26];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    for (int j = 0; j < s[i].size(); j++) {
      if (vol[s[i][j] - 'a'].size() && vol[s[i][j] - 'a'].back() == i) {
        cout << "NO" << endl;
        return 0;
      }
      vol[s[i][j] - 'a'].push_back(i);
    }
  }
  for (int c = 0; c < 26; c++) {
    if (vis[c] || !vol[c].size()) continue;
    for (int i = 0; i < 65; i++) ans[i] = ' ';
    l = 30, r = 30;
    ans[30] = char(c + 'a');
    pos[c] = l;
    Q.push(c);
    vis[c] = true;
    while (!Q.empty()) {
      int now = Q.front();
      Q.pop();
      for (int i = 0; i < vol[now].size(); i++) {
        string tmp = s[vol[now][i]];
        int p = 0;
        for (int j = 0; j < tmp.size(); j++)
          if (tmp[j] == 'a' + now) {
            p = j;
            break;
          }
        for (int j = pos[now] - 1, k = p - 1; k >= 0; l = min(l, j), j--, k--) {
          if (ans[j] != ' ' && ans[j] != tmp[k]) {
            cout << "NO" << endl;
            return 0;
          }
          ans[j] = tmp[k];
          if (pos[ans[j] - 'a'] && pos[ans[j] - 'a'] != j) {
            cout << "NO" << endl;
            return 0;
          }
          pos[ans[j] - 'a'] = j;
        }
        for (int j = pos[now] + 1, k = p + 1; k < tmp.size();
             r = max(r, j), j++, k++) {
          if (ans[j] != ' ' && ans[j] != tmp[k]) {
            cout << "NO" << endl;
            return 0;
          }
          ans[j] = tmp[k];
          if (pos[ans[j] - 'a'] && pos[ans[j] - 'a'] != j) {
            cout << "NO" << endl;
            return 0;
          }
          pos[ans[j] - 'a'] = j;
        }
      }
      for (int i = l; i <= r; i++)
        if (!vis[ans[i] - 'a']) {
          vis[ans[i] - 'a'] = true;
          Q.push(ans[i] - 'a');
        }
    }
    string input;
    for (int i = l; i <= r; i++) input += ans[i];
    final_ans.push_back(input);
  }
  sort(final_ans.begin(), final_ans.end());
  for (int i = 0; i < final_ans.size(); i++) cout << final_ans[i];
  cout << endl;
  return 0;
}
