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 nmax = 1e5 + 42;
int n;
string inp[nmax];
int seen[26];
bool test(int i) {
  memset(seen, 0, sizeof(seen));
  for (auto k : inp[i]) seen[k - 'a']++;
  for (int i = 0; i < 26; i++)
    if (seen[i] > 1) return 0;
  return 1;
}
set<int> nx[26], pr[26];
string out;
bool been[26], there[26];
void dfs(int node) {
  if (been[node]) {
    cout << "NO" << endl;
    exit(0);
  }
  been[node] = 1;
  out = out + char(node + 'a');
  if (nx[node].size()) {
    dfs(*nx[node].begin());
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> inp[i];
  for (int i = 1; i <= n; i++)
    if (test(i) == 0) {
      cout << "NO" << endl;
      return 0;
    }
  for (int i = 1; i <= n; i++) {
    int SZ = inp[i].size();
    for (int j = 1; j < SZ; j++) {
      nx[inp[i][j - 1] - 'a'].insert(inp[i][j] - 'a');
      pr[inp[i][j] - 'a'].insert(inp[i][j - 1] - 'a');
    }
    for (auto k : inp[i]) there[k - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    if (nx[i].size() > 1 || pr[i].size() > 1) {
      cout << "NO" << endl;
      return 0;
    }
  for (int i = 0; i < 26; i++) {
    int pos = i;
    while (1) {
      if (been[pos]) {
        cout << "NO" << endl;
        return 0;
      }
      been[pos] = 1;
      if (nx[pos].size())
        pos = *nx[pos].begin();
      else
        break;
    }
    memset(been, 0, sizeof(been));
  }
  for (int i = 0; i < 26; i++)
    if (nx[i].size() && pr[i].size() == 0) {
      dfs(i);
    } else if (there[i] && nx[i].size() == 0 && pr[i].size() == 0)
      dfs(i);
  cout << out << endl;
  return 0;
}
