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;
long long n, pr[26], dsu[26];
bitset<26> vtd, spc;
long long fd(long long x) {
  if (dsu[x] != x) {
    dsu[x] = fd(dsu[x]);
  }
  return dsu[x];
}
int main() {
  long long i, j, k, l, sz, p;
  string s;
  for (i = 0; i < 26; i++) {
    pr[i] = -1;
    dsu[i] = i;
  }
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    cin >> s;
    sz = s.length();
    for (j = 0; j < sz - 1; j++) {
      k = s[j] - 'a';
      l = s[j + 1] - 'a';
      if (pr[k] == -1) {
        if (fd(k) == fd(l) || vtd[l]) {
          printf("NO\n");
          return 0;
        }
        pr[k] = l;
        dsu[fd(k)] = fd(l);
        vtd[l] = 1;
      } else if (pr[k] != l) {
        printf("NO\n");
        return 0;
      }
    }
    for (j = 0; j < sz; j++) {
      spc[s[j] - 'a'] = 1;
    }
  }
  for (i = 0; i < 26; i++) {
    if (!vtd[i] && spc[i]) {
      for (p = i; p + 1; p = pr[p]) {
        printf("%c", (char)p + 'a');
      }
    }
  }
  printf("\n");
}
