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 nxt[26], prv[26], saw[26], nec[26], used[26];
int n;
void solve() {
  cin >> n;
  for (int i = 0; i < 26; ++i) {
    nxt[i] = prv[i] = -1;
  }
  for (int i = 1; i <= n; ++i) {
    string s;
    cin >> s;
    for (int c = 0; c < 26; ++c) {
      saw[c] = 0;
    }
    for (char c : s) {
      if (saw[c - 'a']) {
        cout << "NO\n";
        return;
      }
      nec[c - 'a'] = 1;
      saw[c - 'a'] = 1;
    }
    for (int j = 0; j + 1 < s.size(); ++j) {
      int a = s[j] - 'a';
      int b = s[j + 1] - 'a';
      if (nxt[a] != -1 && nxt[a] != b) {
        cout << "NO\n";
        return;
      }
      nxt[a] = b;
      if (prv[b] != -1 && prv[b] != a) {
        cout << "NO\n";
        return;
      }
      prv[b] = a;
    }
  }
  string ans;
  for (int i = 0; i < 26; ++i) {
    if (!nec[i]) {
      continue;
    }
    if (prv[i] == -1) {
      for (int v = i; v != -1; v = nxt[v]) {
        if (used[v]) {
          cout << "NO\n";
          return;
        }
        used[v] = 1;
        ans.push_back(v + 'a');
      }
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (nec[i] && !used[i]) {
      cout << "NO\n";
      return;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
