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 = 1e5 + 10;
char str[maxn];
int n, in[30] = {0}, out[30] = {0}, pre[30] = {0}, ne[30] = {0}, book[30] = {0};
bool topu() {
  queue<int> q;
  int i, j, k, h, ans = 0, in1[30] = {0};
  for (i = 1; i <= 26; i++) {
    if (in[i] == 0)
      q.push(i);
    else
      in1[i] = in[i];
  }
  while (q.size()) {
    h = q.front();
    q.pop();
    ans++;
    if (ne[h] == 0) continue;
    in1[ne[h]]--;
    if (in1[ne[h]] <= 0) q.push(ne[h]);
  }
  if (ans < 26)
    return 1;
  else
    return 0;
}
int main() {
  int t, i, j, u, v;
  bool f = 0;
  cin >> n;
  for (i = 0; i < n; i++) {
    scanf("%s", str);
    if (f) continue;
    if (str[1] == 0) {
      u = str[0] - 'a' + 1;
      book[u] = 1;
    }
    for (j = 0; str[j + 1] != 0; j++) {
      u = str[j] - 'a' + 1;
      v = str[j + 1] - 'a' + 1;
      if (book[u]) book[u] = 0;
      if (book[v]) book[v] = 0;
      if (!ne[u] && !pre[v]) {
        ne[u] = v;
        pre[v] = u;
        out[u]++;
        in[v]++;
      } else if (ne[u]) {
        if (ne[u] != v) {
          f = 1;
          break;
        }
      } else if (pre[v]) {
        if (pre[v] != u) {
          f = 1;
          break;
        }
      }
    }
  }
  if (f)
    cout << "NO";
  else {
    if (topu())
      cout << "NO";
    else {
      for (i = 1; i <= 26; i++) {
        if (book[i] && !in[i] && !out[i])
          printf("%c", i + 'a' - 1);
        else if (!in[i] && out[i]) {
          j = i;
          while (ne[j] != 0) {
            printf("%c", j + 'a' - 1);
            j = ne[j];
          }
          printf("%c", j + 'a' - 1);
        }
      }
    }
  }
  return 0;
}
