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;
bool vis[26];
vector<int> mapp[26];
int ma[26][26];
int ingree[26];
int degree[26];
int single[26];
int one[26];
string ans = "";
void dfs(int x) {
  for (int i = 0; i < mapp[x].size(); i++) {
    ans += mapp[x][i] + 'a';
    dfs(mapp[x][i]);
  }
}
bool fan[26];
bool fan2[26];
int main() {
  int n;
  string str;
  int cn = 0;
  while (cin >> n) {
    int f = 1;
    cn = 0;
    memset(single, 0, sizeof(single));
    for (int ii = 1; ii <= n; ii++) {
      cin >> str;
      memset(vis, 0, sizeof(vis));
      for (int i = 0; i < str.size() - 1; i++) {
        if (vis[str[i] - 'a'] == 0) {
          if (ma[str[i] - 'a'][str[i + 1] - 'a'] == 0) {
            mapp[str[i] - 'a'].push_back(str[i + 1] - 'a');
            degree[str[i] - 'a']++;
            ingree[str[i + 1] - 'a']++;
            ma[str[i] - 'a'][str[i + 1] - 'a'] = 1;
          }
          vis[str[i] - 'a'] = 1;
        } else {
          f = 0;
        }
        fan[str[i] - 'a'] = 1;
      }
      fan[str[str.size() - 1] - 'a'] = 1;
      if (vis[str[str.size() - 1]] == 1) {
        f = 0;
      }
      if (str.size() == 1) {
        one[str[0] - 'a'] = 1;
      }
    }
    for (int i = 0; i < 26; i++) {
      if (ingree[i] > 1 || degree[i] > 1) {
        f = 0;
      }
      if (degree[i] == 1 && ingree[i] == 0) single[i] = 1;
      if (one[i] == 1 && (degree[i] == 0 && ingree[i] == 0)) {
        single[i] = 1;
      }
    }
    if (f == 0) {
      cout << "NO" << endl;
    } else {
      for (int i = 0; i < 26; i++) {
        if (single[i] == 1) {
          ans += i + 'a';
          dfs(i);
        }
      }
      for (int i = 0; i < ans.size(); i++) {
        fan2[ans[i] - 'a'] = 1;
      }
      for (int i = 0; i < 26; i++) {
        if (fan[i] != fan2[i]) f = 0;
      }
      if (ans != "" && f)
        cout << ans << endl;
      else {
        cout << "NO" << endl;
      }
    }
  }
  return 0;
}
