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 long long INF = 1e9 + 7;
const int N = 2e2 + 10;
vector<string> v[N];
int nex[N];
int flag[N];
int main() {
  int n;
  cin >> n;
  map<char, int> mc;
  memset((nex), (-1), sizeof(nex));
  while (n--) {
    string str;
    cin >> str;
    int o = 0;
    for (auto &c : str) {
      mc[c] = 1;
      if (flag[c] == 0) flag[c] = 1;
      if (o >> (c - '0') & 1) {
        puts("NO");
        return 0;
      }
      o |= 1 << (c - '0');
    }
    for (int i = 0; i + 1 < str.length(); i++) {
      if (nex[str[i]] == -1) {
        nex[str[i]] = str[i + 1];
        flag[str[i + 1]] = 2;
      }
      if (nex[str[i]] != str[i + 1]) {
        puts("NO");
        return 0;
      }
    }
  }
  string ans;
  for (int i = 'a'; i <= 'z'; i++) {
    if (flag[i] != 1) continue;
    int c = i;
    while (c != -1 && ans.length() <= mc.size()) {
      ans.push_back(c);
      c = nex[c];
    }
  }
  if (ans.length() != mc.size()) {
    puts("NO");
  } else
    puts(ans.c_str());
  return 0;
}
