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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> ne(26, -1);
  set<int> h;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    vector<int> have(26);
    for (auto& x : s) {
      h.insert(x - 'a');
    }
    for (int j = 0; j < (int)s.size(); j++) {
      int k = j + 1;
      if (have[s[j] - 'a']) {
        cout << "NO\n";
        return 0;
      }
      have[s[j] - 'a'] = 1;
      if (k < (int)s.size()) {
        if (ne[s[j] - 'a'] > -1 && ne[s[j] - 'a'] != s[k] - 'a') {
          cout << "NO\n";
          return 0;
        }
        for (int l = 0; l < 26; l++) {
          if (l == s[j] - 'a') {
            continue;
          }
          if (ne[l] == s[k] - 'a') {
            cout << "NO\n";
            return 0;
          }
        }
        ne[s[j] - 'a'] = s[k] - 'a';
      }
    }
  }
  vector<int> ans;
  vector<int> used(26);
  function<bool(int)> dfs = [&](int v) {
    used[v] = 1;
    ans.push_back(v);
    if (ne[v] > -1) {
      if (used[ne[v]] == 1) {
        return false;
      }
      if (used[ne[v]] == 0) {
        if (!dfs(ne[v])) {
          return false;
        }
      }
    }
    used[v] = 2;
    return true;
  };
  for (auto& x : h) {
    if (!used[x]) {
      if (!dfs(x)) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  ans.clear();
  for (auto& x : h) {
    used[x] = 0;
  }
  for (auto& x : h) {
    bool skip = false;
    for (int i = 0; i < 26; i++) {
      if (ne[i] == x) {
        skip = true;
        break;
      }
    }
    if (skip) {
      continue;
    }
    if (!used[x]) {
      if (!dfs(x)) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  for (auto& x : ans) {
    cout << (char)(x + 'a');
  }
  cout << '\n';
  return 0;
}
