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;
const int maxn = 1e5 + 1000;
const int ml = 26;
int n;
string s[maxn];
int nxt[ml], pre[ml], cont[ml];
bool init() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> s[i];
  for (int i = 0; i < ml; ++i) nxt[i] = pre[i] = -1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < ml; ++j) cont[j] = 0;
    for (int j = 1; j < s[i].size(); ++j) {
      int a = int(s[i][j - 1] - 'a');
      int b = int(s[i][j] - 'a');
      if (cont[a]) return false;
      if (nxt[a] != -1 and nxt[a] != b) return false;
      if (pre[b] != -1 and pre[b] != a) return false;
      nxt[a] = b;
      pre[b] = a;
      cont[a]++;
    }
    if (cont[s[i][s[i].size() - 1] - 'a']) return false;
  }
  for (int i = 0; i < ml; ++i) {
    int u = nxt[i];
    while (u != -1 and u != i) u = nxt[u];
    if (u == i) return false;
  }
  for (int i = 0; i < ml; ++i) cont[i] = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < s[i].size(); ++j) {
      cont[s[i][j] - 'a']++;
    }
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  if (init()) {
    for (int i = 0; i < ml; ++i) {
      if (pre[i] == -1 and cont[i]) {
        int u = i;
        while (u != -1) {
          cout << char(u + 'a');
          u = nxt[u];
        }
      }
    }
    cout << '\n';
  } else
    cout << "NO\n";
}
