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;
int n;
string s[100000];
vector<string> ans;
int g[50], mark[50], degIn[50], degOut[50], C[50];
void printNo() {
  cout << "NO" << endl;
  exit(0);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> s[i];
  int flag = 1;
  memset(degOut, 0, sizeof(degOut));
  memset(degIn, 0, sizeof(degIn));
  memset(g, 0, sizeof(g));
  memset(mark, 0, sizeof(mark));
  memset(C, 0, sizeof(C));
  for (int i = 0; i < n; ++i) {
    int len = s[i].length();
    if (len > 26) printNo();
    C[s[i][0] - 'a'] = 1;
    for (int j = 1; j < s[i].length(); ++j) {
      C[s[i][j] - 'a'] = 1;
      if (g[s[i][j - 1] - 'a'] == s[i][j]) continue;
      g[s[i][j - 1] - 'a'] = s[i][j];
      degOut[s[i][j - 1] - 'a']++;
      degIn[s[i][j] - 'a']++;
    }
  }
  for (char c = 'a'; c <= 'z'; ++c) {
    if (degIn[c - 'a'] > 1 || degOut[c - 'a'] > 1) printNo();
  }
  string tmp;
  for (char c = 'a'; c <= 'z'; ++c) {
    if (degIn[c - 'a'] == 0 && C[c - 'a']) {
      char v = c;
      tmp = "";
      do {
        if (mark[v - 'a'] == 0)
          mark[v - 'a'] = 1;
        else
          printNo();
        tmp += v;
        v = g[v - 'a'];
      } while (v);
      ans.push_back(tmp);
    }
  }
  for (int c = 'a'; c <= 'z'; ++c)
    if (C[c - 'a'] && mark[c - 'a'] == 0) printNo();
  sort(ans.begin(), ans.end());
  for (auto v : ans) cout << v;
  cout << endl;
  return 0;
}
