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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return b < a ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int oo = 0x3f3f3f3f;
const int maxn = 100100;
int n;
string a[maxn + 5];
int nxt[26];
bool apr[26];
bool vis[26];
int main() {
  cin >> n;
  memset(nxt, -1, sizeof nxt);
  for (int i = (0), i_end_ = (n); i < i_end_; ++i) {
    cin >> a[i];
    for (int j = (0), j_end_ = ((int((a[i]).size()))); j < j_end_; ++j)
      apr[a[i][j] - 'a'] = 1;
    for (int j = (0), j_end_ = ((int((a[i]).size())) - 1); j < j_end_; ++j) {
      if (~nxt[a[i][j] - 'a']) {
        if (nxt[a[i][j] - 'a'] != a[i][j + 1] - 'a') {
          printf("NO\n");
          return 0;
        } else
          continue;
      }
      if (vis[a[i][j + 1] - 'a']) {
        printf("NO\n");
        return 0;
      }
      vis[nxt[a[i][j] - 'a'] = a[i][j + 1] - 'a'] = 1;
    }
  }
  string ans;
  for (int i = (0), i_end_ = (26); i < i_end_; ++i)
    if (apr[i] && !vis[i]) {
      int u = i;
      while (~u) {
        apr[u] = 0;
        ans += 'a' + u;
        u = nxt[u];
      }
    }
  for (int i = (0), i_end_ = (26); i < i_end_; ++i)
    if (apr[i]) {
      printf("NO\n");
      return 0;
    }
  cout << ans << endl;
  return 0;
}
