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, dd[100005], has[100005], sau[100005], trc[100005], hass[100005];
string ans;
void stop() {
  cout << "NO";
  exit(0);
}
void inkq(int u) {
  while (u != 0) {
    ans.push_back('a' + u - 1);
    has[u] = 1;
    u = sau[u];
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < s.size() - 1; j++) {
      int u = s[j] - 'a' + 1;
      int v = s[j + 1] - 'a' + 1;
      if (trc[v] && trc[v] != u) stop();
      if (sau[u] && sau[u] != v) stop();
      if (sau[v] && sau[v] == u) stop();
      if (trc[u] && trc[u] == v) stop();
      sau[u] = v;
      trc[v] = u;
    }
    if (s.size() == 1) dd[s[0] - 'a' + 1] = 1;
    for (int j = 0; j < s.size(); j++) hass[s[j] - 'a' + 1] = 1;
  }
  bool ok = 0;
  for (int i = 1; i <= 'z' - 'a' + 1; i++) {
    if (has[i]) continue;
    if ((trc[i] == 0) && (sau[i] || dd[i])) {
      ok = 1;
      inkq(i);
    }
  }
  if (!ok) stop();
  for (int i = 1; i <= 'z' - 'a' + 1; i++)
    if (has[i] != hass[i]) stop();
  cout << ans;
}
