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 = 2e5 + 10;
char str[maxn];
int vis[maxn], out[maxn], in[maxn], input[maxn];
vector<string> vec;
vector<int> G[maxn];
string temp;
int ok;
void dfs(int cur) {
  if (ok == 0) return;
  vis[cur] = 1;
  temp += 'a' + cur;
  for (auto v : G[cur]) {
    if (vis[v]) ok = 0;
    dfs(v);
  }
}
bool solve() {
  for (int i = 0; i < 26; i++) {
    out[i] += G[i].size();
    for (auto o : G[i]) in[o]++;
  }
  for (int i = 0; i < 26; i++) {
    if (in[i] > 1 || out[i] > 1) return false;
  }
  for (int i = 0; i < 26; i++) {
    ok = 1;
    temp = "";
    if (in[i] == 0 && out[i]) dfs(i);
    vec.push_back(temp);
    if (ok == 0) {
      return false;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (vis[i] == 0 && in[i]) return false;
    if (vis[i] == 0 && input[i]) {
      string t = "";
      t += ('a' + i);
      vec.push_back(t);
    }
  }
  sort(vec.begin(), vec.end());
  string ans = "";
  for (auto o : vec) ans += o;
  cout << ans << endl;
  return true;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> str;
    int len = strlen(str);
    if (len == 1) {
      input[str[0] - 'a'] = 1;
    } else {
      for (int j = 0; j < len - 1; j++) {
        G[str[j] - 'a'].push_back(str[j + 1] - 'a');
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    sort(G[i].begin(), G[i].end());
    G[i].resize(unique(G[i].begin(), G[i].end()) - G[i].begin());
  }
  if (solve() == false) puts("NO");
  return 0;
}
