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>
#pragma GCC optimize("O3")
const double PI = acos(-1);
using namespace std;
const int MAX = 3 * 100 * 1000 + 10;
char buffer[MAX];
string S[MAX];
int n;
int cnt[128];
int prv[128], nxt[128];
bool isSet(string& s) {
  set<char> S{s.begin(), s.end()};
  return ((int)S.size()) == ((int)s.size());
}
void compress() {
  sort(S, S + n, [](const string& a, const string& b) {
    return ((int)a.size()) < ((int)b.size());
  });
  set<string> aux;
  for (int t = 0; t < (n); t++) {
    string s = S[t];
    for (int i = 0; i < ((int)s.size()); i++) {
      string u;
      for (int j = i; j < ((int)s.size()); j++) {
        u += s[j];
        aux.erase(u);
      }
    }
    aux.insert(s);
  }
  copy(aux.begin(), aux.end(), S);
  n = ((int)aux.size());
}
string solve() {
  for (int i = 0; i < (n); i++) {
    if (!isSet(S[i])) return "NO";
  }
  compress();
  int ctr = 0;
  for (int i = 0; i < (n); i++) {
    for (char c : S[i]) {
      int x = c;
      ctr += !cnt[x];
      cnt[x]++;
    }
  }
  for (int i = 0; i < (n); i++) {
    string s = S[i];
    for (int j = 0; j + 1 < ((int)s.size()); j++) {
      int x = s[j], y = s[j + 1];
      if (nxt[x] && nxt[x] != y) return "NO";
      if (prv[y] && prv[y] != x) return "NO";
      nxt[x] = y;
      prv[y] = x;
    }
  }
  int m = 0;
  for (int i = 'a'; i <= 'z'; i++)
    if (cnt[i]) {
      if (!prv[i]) {
        int cur = i;
        while (cur) {
          buffer[m++] = cur;
          cur = nxt[cur];
        }
      }
    }
  buffer[m] = 0;
  if (m != ctr) return "NO";
  return string(buffer);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < (n); i++) {
    scanf("%s", buffer);
    S[i] = string(buffer);
  }
  puts(solve().c_str());
  return 0;
}
