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;
bool check = true, visited[26];
int to[26], deg[26], father[26], cnt = 0;
char str[200002], ans[200002];
int find(int x) { return father[x] != x ? father[x] = find(father[x]) : x; }
void go(int s) {
  for (int now = s; now != -1; now = to[now])
    ans[cnt++] = (char)((int)'a' + now);
}
void merge(int x, int y) {
  x = find(x), y = find(y);
  if (x == y) check = false;
  father[x] = y;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < 26; i++)
    visited[i] = false, deg[i] = 0, to[i] = -1, father[i] = i;
  for (int i = 0; i < n; i++) {
    scanf("%s", str);
    for (int i = 0; str[i]; i++) visited[(int)(str[i] - 'a')] = true;
    for (int i = 0; str[i + 1]; i++) {
      int s = (int)(str[i] - 'a');
      int t = to[s], next = (int)(str[i + 1] - 'a');
      if (t >= 0 && t != next) check = false;
      if (t != next) merge(s, next);
      to[s] = next;
    }
  }
  for (int i = 0; i < 26; i++) deg[to[i]]++;
  for (int i = 0; i < 26; i++)
    if (deg[i] > 1) check = false;
  if (!check)
    printf("NO\n");
  else {
    for (int i = 0; i < 26; i++)
      if (visited[i] && !deg[i]) go(i);
    if (!check)
      printf("NO\n");
    else
      for (int i = 0; i < cnt; i++) printf("%c", ans[i]);
  }
  return 0;
}
