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 max_n = 100111, max_c = 30, inf = 1011111111;
int n, used[2][max_n][max_c];
vector<int> g[max_c];
char s[max_n];
string cur, all[2][max_n];
int q1 = 0, q2 = 1;
void fail() {
  printf("NO\n");
  exit(0);
}
void build_used() {
  memset(used[q1], -1, sizeof(used[q1]));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < all[q1][i].length(); ++j) {
      if (used[q1][i][all[q1][i][j] - 'a'] != -1) {
        fail();
      }
      used[q1][i][all[q1][i][j] - 'a'] = j;
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    all[0][i] = s;
  }
  build_used();
  for (int i = 0; i < 26; ++i) {
    int m = 0;
    string cur = "";
    for (int j = 0; j < n; ++j) {
      if (used[q1][j][i] != -1) {
        if (cur == "") {
          cur = all[q1][j];
        } else {
          int pos = cur.find(i + 'a');
          string ncur = cur;
          if (pos < used[q1][j][i]) {
            ncur = all[q1][j].substr(0, used[q1][j][i] - pos) + cur;
          }
          if (cur.length() - pos < all[q1][j].length() - used[q1][j][i]) {
            int st = used[q1][j][i] + cur.length() - pos;
            ncur += all[q1][j].substr(st);
          }
          cur.swap(ncur);
          pos = cur.find(i + 'a');
          if (cur.substr(pos - used[q1][j][i], all[q1][j].length()) !=
              all[q1][j]) {
            fail();
          }
        }
      }
    }
    for (int j = 0; j < n; ++j) {
      if (used[q1][j][i] == -1) {
        all[q2][m] = all[q1][j];
        ++m;
      }
    }
    if (cur != "") {
      all[q2][m] = cur;
      ++m;
    }
    swap(q1, q2);
    swap(n, m);
    build_used();
  }
  sort(all[q1], all[q1] + n);
  for (int i = 0; i < n; ++i) {
    printf("%s", all[q1][i].c_str());
  }
  printf("\n");
  return 0;
}
