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;
vector<int> G[30];
int GG[30][30];
char str[100005];
vector<int> ans;
int vis[30];
bool dfs(int x) {
  vis[x] = 1;
  ans.push_back(x);
  for (auto i : G[x]) {
    if (vis[i] || !dfs(i)) {
      return false;
    }
  }
  return true;
}
int d[30];
int oc[30];
int main(int argc, const char* argv[]) {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", str);
    int l = (int)strlen(str);
    for (int j = 0; j < l; j++) {
      oc[str[j] - 'a'] = 1;
      if (j == l - 1) {
        break;
      }
      if (!GG[str[j] - 'a'][str[j + 1] - 'a']) {
        G[str[j] - 'a'].push_back(str[j + 1] - 'a');
        d[str[j + 1] - 'a']++;
        GG[str[j] - 'a'][str[j + 1] - 'a'] = 1;
      }
    }
  }
  memset(vis, 0, sizeof vis);
  for (int i = 0; i < 26; i++) {
    if (!oc[i]) continue;
    if (!vis[i] && d[i] == 0 && !dfs(i)) {
      puts("NO");
      return 0;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (G[i].size() > 1) {
      puts("NO");
      return 0;
    }
    if (G[i].size() && vis[i] == 0) {
      puts("NO");
      return 0;
    }
  }
  for (int i = 0; i < ans.size(); i++) {
    putchar('a' + ans[i]);
  }
  putchar('\n');
  return 0;
}
