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;
using lli = long long int;
using ll = lli;
using llu = unsigned long long int;
using pii = tuple<lli, lli>;
using piii = tuple<lli, lli, lli>;
using vi = vector<lli>;
using vii = vector<pii>;
using viii = vector<piii>;
using vvi = vector<vi>;
using vvii = vector<vii>;
using vviii = vector<viii>;
using vb = vector<bool>;
using vvb = vector<vb>;
const int MAXN = 100 * 1000;
string s[MAXN];
char tmp[MAXN + 1];
int main() {
  int n;
  scanf("%d", &n);
  for (lli i = 0; i < (lli)(n); ++i) {
    scanf("%s", tmp);
    s[i] = tmp;
  }
  vi nxt(26, -1);
  vi pres(26);
  for (lli i = 0; i < (lli)(n); ++i) {
    for (auto c : s[i]) pres[c - 'a'] = true;
    for (lli j = 0; j < (lli)((ll)s[i].size() - 1); ++j) {
      int c = s[i][j] - 'a';
      int v = s[i][j + 1] - 'a';
      if (nxt[c] != -1 && nxt[c] != v) {
        puts("NO");
        return 0;
      }
      nxt[c] = v;
    }
  }
  vi indeg(26), seen(26);
  for (lli c = 0; c < (lli)(26); ++c)
    if (nxt[c] != -1) indeg[nxt[c]]++;
  string ans;
  while (true) {
    int u = -1;
    for (lli c = 0; c < (lli)(26); ++c)
      if (indeg[c] == 0 && pres[c] && !seen[c]) {
        u = c;
        break;
      }
    if (u == -1) break;
    while (u != -1) {
      if (seen[u]) {
        puts("NO");
        return 0;
      }
      ans += u + 'a';
      seen[u] = 1;
      u = nxt[u];
    }
  }
  for (lli c = 0; c < (lli)(26); ++c)
    if (pres[c] && !seen[c]) {
      puts("NO");
      return 0;
    }
  printf("%s\n", ans.c_str());
}
