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 maxn = 1e5 + 100;
const int L = 26;
char s[maxn];
int nex[30], pre[30];
bool vis[30];
int main() {
  int n;
  for (; scanf("%d", &n) != EOF;) {
    for (int i = (0), _i = (L - 1); i <= _i; ++i) nex[i] = pre[i] = -1;
    for (int i = (0), _i = (L - 1); i <= _i; ++i) vis[i] = 0;
    bool flag = 1;
    for (int _ = (1), __ = (n); _ <= __; ++_) {
      scanf("%s", s);
      int len = strlen(s);
      for (int i = (0), _i = (len - 1); i <= _i; ++i) vis[s[i] - 'a'] = 1;
      for (int i = (1), _i = (len - 1); i <= _i; ++i) {
        int a = s[i - 1] - 'a', b = s[i] - 'a';
        if ((nex[a] != -1 && nex[a] != b) || (pre[b] != -1 && pre[b] != a)) {
          flag = 0;
        }
        nex[a] = b;
        pre[b] = a;
      }
    }
    if (!flag)
      puts("NO");
    else {
      string ans = "";
      for (int i = (0), _i = (L - 1); i <= _i; ++i)
        if (vis[i] && pre[i] == -1) {
          for (int x = i; x != -1; x = nex[x])
            vis[x] = 0, ans += (char)(x + 'a');
        }
      for (int i = (0), _i = (L - 1); i <= _i; ++i)
        if (vis[i]) {
          flag = 0;
          break;
        }
      if (!flag)
        puts("NO");
      else
        cout << ans << endl;
    }
  }
  return 0;
}
