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 prevc[26], nextc[26];
bool isE[26];
char s[100001];
bool read_data() {
  int n;
  memset(prevc, 0xff, sizeof(prevc));
  memset(nextc, 0xff, sizeof(nextc));
  memset(isE, 0, sizeof(isE));
  cin >> n;
  while (n--) {
    scanf(" %s", s);
    int len = strlen(s);
    for (int i = 0; i < len; i++) {
      int c = (s[i] - 'a');
      isE[c] = true;
      if (i != 0) {
        if (prevc[c] == -1 || prevc[c] == (s[i - 1] - 'a'))
          prevc[c] = (s[i - 1] - 'a');
        else
          return false;
      }
      if (i != len - 1) {
        if (nextc[c] == -1 || nextc[c] == (s[i + 1] - 'a'))
          nextc[c] = (s[i + 1] - 'a');
        else
          return false;
      }
    }
  }
  return true;
}
bool valid() {
  vector<string> s;
  s.assign(26, "");
  vector<int> isv;
  isv.assign(26, -1);
  queue<int> q;
  for (int i = 0; i < 26; i++)
    if (prevc[i] == -1 && isE[i]) {
      isv[i] = i;
      q.push(i);
      char ci = 'a' + i;
      s[i].push_back(ci);
    }
  while (!q.empty()) {
    int c = q.front();
    q.pop();
    int nc = nextc[c];
    if (nc == -1) continue;
    if (isv[nc] != -1) return false;
    isv[nc] = isv[c];
    q.push(nc);
    char nci = 'a' + nc;
    s[isv[c]].push_back(nci);
  }
  for (int i = 0; i < 26; i++)
    if (isE[i] & isv[i] == -1) return false;
  sort(s.begin(), s.end());
  for (int i = 0; i < 26; i++) cout << s[i];
  cout << endl;
  return true;
}
int main() {
  if (!read_data() || !valid()) {
    cout << "NO" << endl;
  }
  return 0;
}
