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 mod = 1000000007;
int nxt[26];
int pr[26];
bool used[26];
bool was[26];
string t = "";
void dfs(int c) {
  if (used[c]) {
    cout << "NO\n";
    exit(0);
  }
  t += c + 'a';
  used[c] = true;
  if (nxt[c] != -1) {
    dfs(nxt[c]);
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<string> s(n);
  for (int i = 0; i < (int)(26); ++i) {
    nxt[i] = -1;
    pr[i] = -1;
  }
  for (int i = 0; i < (int)(n); ++i) {
    cin >> s[i];
    for (int j = 0; j < (int)(s[i].size()); ++j) {
      for (int k = 0; k < (int)(j); ++k) {
        if (s[i][j] == s[i][k]) {
          cout << "NO\n";
          return 0;
        }
      }
      was[s[i][j] - 'a'] = true;
    }
    for (int j = 0; j < (int)(s[i].size() - 1); ++j) {
      int c1 = s[i][j] - 'a';
      int c2 = s[i][j + 1] - 'a';
      if (nxt[c1] != -1 && nxt[c1] != c2) {
        cout << "NO\n";
        return 0;
      }
      nxt[c1] = c2;
      if (pr[c2] != -1 && pr[c2] != c1) {
        cout << "NO\n";
        return 0;
      }
      pr[c2] = c1;
    }
  }
  for (int i = 0; i < (int)(26); ++i) {
    if (was[i] && !used[i] && pr[i] == -1) {
      dfs(i);
    }
  }
  for (int i = 0; i < (int)(26); ++i) {
    if (was[i] && !used[i]) {
      cout << "NO\n";
      return 0;
    }
  }
  cout << t << endl;
  return 0;
}
