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, m, la[26], lb[26];
bool o[26], vis[100005], v2[26];
char x[500005];
vector<string> v, ans;
vector<pair<int, int> > to[100005], cur;
void dfs(int a, int b) {
  cur.push_back(make_pair(b, a));
  vis[a] = 1;
  for (int i = 0; i < to[a].size(); i++)
    if (!vis[to[a][i].first]) dfs(to[a][i].first, b + to[a][i].second);
}
bool cmp(const string &a, const string &b) { return a + b < b + a; }
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", x);
    v.push_back(string(x));
    m = strlen(x);
    for (int j = 0; j < m; j++) {
      x[j] -= 'a';
      if (o[x[j]]) {
        to[la[x[j]]].push_back(make_pair(i, lb[x[j]] - j));
        to[i].push_back(make_pair(la[x[j]], j - lb[x[j]]));
      } else {
        o[x[j]] = 1;
        la[x[j]] = i;
        lb[x[j]] = j;
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (!vis[i]) {
      cur.clear();
      dfs(i, 0);
      sort(cur.begin(), cur.end());
      for (int j = cur.size() - 1; j >= 0; j--) cur[j].first -= cur[0].first;
      m = 0;
      for (int j = 0; j < cur.size(); j++) {
        m = max(m, cur[j].first + (int)v[cur[j].second].size());
        memcpy(x + cur[j].first, v[cur[j].second].c_str(),
               v[cur[j].second].size());
      }
      x[m] = 0;
      for (int j = 0; j < cur.size(); j++)
        if (memcmp(x + cur[j].first, v[cur[j].second].c_str(),
                   v[cur[j].second].size())) {
          printf("NO\n");
          return 0;
        }
      ans.push_back(string(x));
    }
  for (int i = 0; i < ans.size(); i++)
    for (int j = 0; j < ans[i].size(); j++)
      if (v2[ans[i][j] - 'a']) {
        printf("NO\n");
        return 0;
      } else
        v2[ans[i][j] - 'a'] = 1;
  sort(ans.begin(), ans.end(), cmp);
  for (int i = 0; i < ans.size(); i++) printf("%s", ans[i].c_str());
  printf("\n");
  return 0;
}
