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;
int n, in[26];
bool F[26][26], kt, flag[26], appear[26];
string A[100010];
vector<string> res;
queue<int> Q;
void bfs(int x) {
  string kq = "";
  while (!Q.empty()) Q.pop();
  flag[x] = false;
  Q.push(x);
  while (!Q.empty()) {
    int u = Q.front(), cnt = 0, tmp;
    Q.pop();
    kq.push_back(u + 'a');
    for (int i = 0; i <= 25; ++i)
      if (F[u][i]) ++cnt, tmp = i;
    if (!cnt) continue;
    if (cnt > 1) {
      kt = false;
      return;
    }
    flag[tmp] = false;
    Q.push(tmp);
  }
  res.push_back(kq);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> A[i];
  memset(F, false, sizeof(F));
  memset(flag, true, sizeof(flag));
  memset(in, 0, sizeof(in));
  memset(appear, false, sizeof(appear));
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= A[i].size() - 1; ++j)
      if (!F[A[i][j - 1] - 'a'][A[i][j] - 'a'])
        F[A[i][j - 1] - 'a'][A[i][j] - 'a'] = true, ++in[A[i][j] - 'a'];
    for (int j = 0; j <= A[i].size() - 1; ++j) appear[A[i][j] - 'a'] = true;
  }
  kt = true;
  for (int i = 0; i <= 25; ++i)
    if (in[i] > 1) kt = false;
  for (int i = 0; i <= 25; ++i)
    if (!in[i] && appear[i] && kt) bfs(i);
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= A[i].size() - 1; ++j)
      if (flag[A[i][j] - 'a']) {
        kt = false;
        break;
      }
  if (!kt || res.empty()) {
    cout << "NO";
    return 0;
  }
  sort(res.begin(), res.end());
  for (int i = 0; i <= res.size() - 1; ++i) cout << res[i];
  return 0;
}
