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;
vector<string> s;
bool testbit(int n, int m) { return (n >> m) & 1; }
int get_any(int n) {
  if (n == 0) return -1;
  for (int i = 0;; i++)
    if (testbit(n, i)) return i;
  return -1;
}
string solve() {
  vector<int> mask;
  for (int i = 0; i < n; i++) {
    int curmask = 0;
    for (int j = 0; j < (int)s[i].size(); j++) {
      if (testbit(curmask, s[i][j] - 'a'))
        return "NO";
      else
        curmask |= 1 << (s[i][j] - 'a');
    }
    mask.push_back(curmask);
  }
  for (int i = 0; i < n; i++) {
    int curmask = mask[i];
    for (int j = i + 1; j < n; j++)
      if (curmask & mask[j]) {
        int same = get_any(curmask & mask[j]);
        int p1 = -1;
        int p2 = -1;
        for (int k = 0; k < s[i].size(); k++)
          if (s[i][k] - 'a' == same) p1 = k;
        for (int k = 0; k < s[j].size(); k++)
          if (s[j][k] - 'a' == same) p2 = k;
        if (p1 < p2) {
          swap(p1, p2);
          swap(s[i], s[j]);
          swap(mask[i], mask[j]);
        }
        for (int k = 0; k < s[j].size(); k++) {
          assert(0 <= p1 - p2 + k);
          if (p1 - p2 + k < s[i].size()) {
            if (s[i][p1 - p2 + k] != s[j][k]) return "NO";
          } else
            s[i].push_back(s[j][k]);
        }
        mask.erase(mask.begin() + j);
        s.erase(s.begin() + j);
        j = i;
        n--;
        curmask = 0;
        for (int k = 0; k < (int)s[i].size(); k++) {
          if (testbit(curmask, s[i][k] - 'a'))
            return "NO";
          else
            curmask |= 1 << (s[i][k] - 'a');
        }
      }
  }
  sort(s.begin(), s.end());
  return accumulate(s.begin(), s.end(), string(""));
}
int main() {
  scanf("%d", &n);
  string res = "NO";
  if (n < 700) {
    s.resize(n);
    for (int i = 0; i < n; i++) cin >> s[i];
    res = solve();
  }
  cout << res << endl;
  return 0;
}
