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 + 100;
string s;
int p[30], root[30];
bool used[30];
void dfs(int v) {
  s += v + 'a';
  if (used[v]) {
    puts("NO");
    exit(0);
  }
  used[v] = 1;
  if (p[v] == v) {
    return;
  }
  dfs(p[v]);
}
int main() {
  for (int i = 0; i < 30; i++) {
    p[i] = i;
    root[i] = 1;
  }
  set<int> alp;
  int n;
  scanf("%d\n", &n);
  for (int i = 1; i <= n; i++) {
    char s[N];
    scanf("%s", s);
    for (int j = 0; j < strlen(s); j++) {
      alp.insert(s[j] - 'a');
    }
    for (int j = 0; j < strlen(s) - 1; j++) {
      if ((p[s[j] - 'a'] != s[j] - 'a' && p[s[j] - 'a'] != s[j + 1] - 'a') ||
          s[j] - 'a' == s[j + 1] - 'a') {
        return puts("NO"), 0;
      }
      p[s[j] - 'a'] = s[j + 1] - 'a';
      root[s[j + 1] - 'a'] = 0;
    }
  }
  for (int i = 0; i < 26; i++) {
    for (int j = 0; j < 26; j++) {
      if (i != j && ((p[i] == p[j] && i != p[i] && j != p[j]) ||
                     (p[i] == j && p[j] == i))) {
        return puts("NO"), 0;
      }
    }
  }
  vector<string> ans;
  for (int i = 0; i < 26; i++) {
    if (alp.count(i) && !used[i] && root[i]) {
      dfs(i);
      ans.push_back(s);
      s.clear();
    }
  }
  if (ans.empty()) {
    return puts("NO"), 0;
  }
  sort(ans.begin(), ans.end());
  string str;
  for (int i = 0; i < ans.size(); i++) {
    str += ans[i];
  }
  for (int i = 0; i < str.size(); i++) {
    alp.erase(str[i] - 'a');
  }
  if (!alp.empty()) {
    return puts("NO"), 0;
  }
  printf("%s", str.c_str());
}
