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;
long long pwr(long long base, long long p, long long mod = (1000000007LL)) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
set<int> adj[30];
bool done[30], instack[30], present[30];
int indeg[30];
stack<int> st;
void dfs(int v) {
  done[v] = instack[v] = true;
  st.push(v);
  for (auto vv : adj[v])
    if (!done[vv])
      dfs(vv);
    else if (instack[vv]) {
      cout << "NO";
      exit(0);
    }
  st.pop();
  instack[v] = false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  while (n--) {
    string str;
    cin >> str;
    for (int i = 0; i < (int)str.length() - 1; i++)
      adj[str[i] - 'a'].insert(str[i + 1] - 'a');
    for (int i = 0; i < (int)str.length(); i++) present[str[i] - 'a'] = true;
  }
  for (int i = 0; i < 26; i++)
    if (present[i] && !done[i]) dfs(i);
  for (int i = 0; i < 26; i++)
    for (auto it : adj[i]) {
      indeg[it]++;
      if (it == i || indeg[it] == 2 || (int)adj[i].size() > 1) {
        cout << "NO";
        return 0;
      }
    }
  vector<string> ans;
  for (int i = 0; i < 26; i++)
    if (present[i] && indeg[i] == 0) {
      string temp = "";
      int v = i;
      while (!adj[v].empty()) {
        temp += ((char)('a' + v));
        v = (*adj[v].begin());
      }
      temp += ((char)('a' + v));
      ans.push_back(temp);
    }
  sort(ans.begin(), ans.end());
  for (auto it : ans) cout << it;
  return 0;
}
