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 N = 1e5 + 5;
int n, pre[26], nxt[26], fre[26], vis[26];
char str[N];
void solve() {
  scanf("%d", &n);
  memset(pre, -1, sizeof(pre));
  memset(nxt, -1, sizeof(nxt));
  for (int i = 0; i < n; i++) {
    memset(fre, 0, sizeof(fre));
    scanf("%s", str);
    int sz = strlen(str);
    for (int j = 0; j < sz; j++) {
      fre[str[j] - 'a']++;
      vis[str[j] - 'a'] = 1;
      if (fre[str[j] - 'a'] > 1) {
        puts("NO");
        return;
      }
    }
    for (int j = 0; j < sz; j++) {
      if (j) {
        if (pre[str[j] - 'a'] != -1 && pre[str[j] - 'a'] != str[j - 1] - 'a') {
          puts("NO");
          return;
        } else
          pre[str[j] - 'a'] = str[j - 1] - 'a';
      }
      if (j < sz - 1) {
        if (nxt[str[j] - 'a'] != -1 && nxt[str[j] - 'a'] != str[j + 1] - 'a') {
          puts("NO");
          return;
        } else
          nxt[str[j] - 'a'] = str[j + 1] - 'a';
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    int cur = nxt[i];
    while (cur != -1) {
      if (cur == i) {
        puts("NO");
        return;
      }
      cur = nxt[cur];
    }
  }
  for (int i = 0; i < 26; i++) {
    if (pre[i] == -1 && vis[i]) {
      int cur = i;
      while (cur != -1) {
        printf("%c", cur + 'a');
        cur = nxt[cur];
      }
    }
  }
  puts("");
}
int main() {
  solve();
  return 0;
}
