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 namespace placeholders;
template <class T>
void mini(T &l, T r) {
  l = min(l, r);
}
template <class T>
void maxi(T &l, T r) {
  l = max(l, r);
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const int S = 26, N = 1E5 + 10;
int nex[S], pre[S];
int vis[S];
char s[N];
int main() {
  int n;
  scanf("%d", &n);
  fill_n(nex, S, -1);
  fill_n(pre, S, -1);
  bool flag = true;
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    set<char> lst;
    lst.insert(s[0]), vis[s[0] - 'a'] = true;
    for (int i = 1; s[i]; ++i) {
      int u = s[i - 1] - 'a', v = s[i] - 'a';
      flag &= lst.find(s[i]) == lst.end();
      lst.insert(s[i]), vis[s[i] - 'a'] = true;
      if (~nex[u] && nex[u] != v || ~pre[v] && pre[v] != u) flag = false;
      nex[u] = v, pre[v] = u;
    }
  }
  if (flag) {
    string ans;
    int len = 0;
    for (int i = 0; i < S; ++i) {
      if (vis[i] && !~pre[i]) {
        int u = i;
        for (; ~nex[u]; u = nex[u]) ans.push_back('a' + u);
        ans.push_back(u + 'a');
      }
      len += vis[i];
    }
    puts(ans.size() != len ? "NO" : ans.c_str());
  } else {
    puts("NO");
  }
  return 0;
}
