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 nxt[30], prv[30];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  vector<int> nodes;
  int n;
  cin >> n;
  memset(nxt, -1, sizeof nxt);
  memset(prv, -1, sizeof prv);
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int j = 0; j < s.length(); j++) {
      nodes.push_back(s[j] - 'a');
      if (j) {
        int x = s[j - 1] - 'a', y = s[j] - 'a';
        if (nxt[x] != -1 && nxt[x] != y) return puts("NO"), 0;
        if (prv[y] != -1 && prv[y] != x) return puts("NO"), 0;
        nxt[x] = y;
        prv[y] = x;
      }
    }
  }
  sort(nodes.begin(), nodes.end());
  nodes.erase(unique(nodes.begin(), nodes.end()), nodes.end());
  for (int i = 0; i < nodes.size(); i++) {
    int x = nodes[i];
    for (int y = nxt[x]; y != -1; y = nxt[y])
      if (y == x) return puts("NO"), 0;
  }
  string ans = "";
  for (int i = 0; i < nodes.size(); i++) {
    int x = nodes[i];
    if (prv[x] != -1) continue;
    for (int y = x; y != -1; y = nxt[y]) ans += 'a' + y;
  }
  cout << ans << endl;
}
