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;
struct List {
  int p1, p2;
};
int n;
string s[100005];
int mark[100];
int pos[30];
bool flag;
int Merge(int x, int y, int p1, int p2) {
  int pp = p1;
  for (int i = 0; i <= min(p1, p2); i++)
    if (s[x][p1 - i] != s[y][p2 - i]) {
      flag = 1;
      break;
    }
  if (flag) return 0;
  int l1 = s[x].length(), l2 = s[y].length();
  for (int i = 1; i <= min(l1 - p1 - 1, l2 - p2 - 1); i++)
    if (s[x][p1 + i] != s[y][p2 + i]) {
      flag = 1;
      break;
    }
  if (flag) return 0;
  string head = "", tail = "";
  if (p1 < p2)
    for (int i = 0; i < p2 - p1; i++) head = head + s[y][i];
  if (l1 - p1 < l2 - p2)
    for (int i = p2 + l1 - p1; i < l2; i++) tail = tail + s[y][i];
  s[x] = head + s[x] + tail;
  s[y] = "";
  for (int i = 0; i < s[x].length(); i++) {
    if (mark[s[x][i] - 'a'] == x || mark[s[x][i] - 'a'] == y) {
      mark[s[x][i] - 'a'] = x;
      pos[s[x][i] - 'a'] = i;
    }
  }
  return head.length() + p1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    for (int j = 0; j < s[i].length(); j++) {
      mark[s[i][j] - 'a']++;
      if (mark[s[i][j] - 'a'] > 1) {
        flag = 1;
        break;
      }
    }
    memset(mark, 0, sizeof(mark));
    if (flag) break;
  }
  if (flag) {
    printf("NO");
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < s[i].length(); j++) {
      if (!mark[s[i][j] - 'a']) {
        mark[s[i][j] - 'a'] = i;
        pos[s[i][j] - 'a'] = j;
      } else if (mark[s[i][j] - 'a'] != i) {
        int temp;
        temp = Merge(i, mark[s[i][j] - 'a'], j, pos[s[i][j] - 'a']);
        j = 0;
      }
      if (flag) break;
    }
  }
  if (flag) {
    printf("NO");
    return 0;
  }
  sort(s + 1, s + n + 1);
  memset(mark, 0, sizeof(mark));
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < s[i].length(); j++) {
      mark[s[i][j] - 'a']++;
      if (mark[s[i][j] - 'a'] > 1) {
        flag = 1;
        break;
      }
    }
    if (flag) break;
  }
  if (flag) {
    printf("NO");
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (s[i] != "") cout << s[i];
  return 0;
}
