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>
#pragma comment(linker, "/STACK:250000000")
using namespace std;
int n;
char buf[100000 + 1];
int cnt[26];
set<int> G[26];
set<int> Gt[26];
int color[26];
int used[26];
bool cycle(int v) {
  color[v] = 1;
  for (int to : G[v]) {
    if (color[to] == 1) return 1;
    if (!color[to] && cycle(to)) return 1;
  }
  color[v] = 2;
  return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", buf);
    memset(cnt, 0, sizeof(cnt));
    int m = strlen(buf);
    for (int j = 0; j < m; ++j) {
      cnt[buf[j] - 'a']++;
      used[buf[j] - 'a'] = 1;
    }
    bool ok = 1;
    for (int j = 0; j < 26; ++j) ok &= cnt[j] <= 1;
    if (!ok) {
      printf("NO\n");
      return 0;
    }
    for (int j = 0; j < m - 1; ++j) {
      int from = buf[j] - 'a';
      int to = buf[j + 1] - 'a';
      G[from].insert(to);
      Gt[to].insert(from);
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (G[i].size() > 1 || Gt[i].size() > 1) {
      printf("NO\n");
      return 0;
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (!color[i] && cycle(i)) {
      printf("NO\n");
      return 0;
    }
  }
  vector<string> res;
  for (int i = 0; i < 26; ++i) {
    if (!used[i]) continue;
    if (!Gt[i].size()) {
      int v = i;
      string add = "";
      add += (char)(v + 'a');
      while (G[v].size()) {
        int to = *G[v].begin();
        add += (char)(to + 'a');
        v = to;
      }
      res.push_back(add);
    }
  }
  sort(res.begin(), res.end());
  for (string& s : res) printf("%s", s.c_str());
  printf("\n");
  return 0;
}
