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;
int nxt[256];
bool u[256];
int n;
int pre[256];
vector<char> ans;
inline void dfs(int v) {
  if (u[v]) {
    cout << "NO";
    exit(0);
  }
  u[v] = 1;
  ans.emplace_back(char(v));
  if (nxt[v]) dfs(nxt[v]);
}
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  while (n--) {
    string s;
    cin >> s;
    for (int i = 0; i + 1 < s.size(); ++i) {
      if (nxt[s[i]]) {
        if (nxt[s[i]] != s[i + 1]) {
          cout << "NO";
          return 0;
        }
      } else {
        nxt[s[i]] = s[i + 1];
      }
    }
    int cnt[256];
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i < s.size(); ++i) {
      u[s[i]] = 1;
      if (++cnt[s[i]] > 1) {
        cout << "NO";
        return 0;
      }
    }
  }
  for (int i = 0; i < 256; ++i) u[i] ^= 1;
  for (int i = 0; i < 256; ++i) {
    pre[nxt[i]] = 1;
  }
  for (int i = 0; i < 256; ++i) {
    if (!pre[i] && !u[i]) dfs(i);
  }
  for (int i = 0; i < 256; ++i) {
    if (!u[i]) {
      cout << "NO";
      return 0;
    }
  }
  for (auto it : ans) cout << it;
  return 0;
}
