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;
vector<int> ans;
int n, a[100005], pre[30], nxt[30];
bool vis[30];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    int m = 0;
    char c = getchar();
    while (c < 'a' || c > 'z') c = getchar();
    while (c >= 'a' && c <= 'z') {
      a[++m] = c - 'a' + 1;
      vis[a[m]] = 1;
      c = getchar();
    }
    for (int j = 1; j < m; ++j) {
      if ((pre[a[j + 1]] && pre[a[j + 1]] != a[j]) ||
          (nxt[a[j]] && nxt[a[j]] != a[j + 1])) {
        puts("NO");
        return 0;
      }
      pre[a[j + 1]] = a[j], nxt[a[j]] = a[j + 1];
    }
  }
  for (int i = 1; i <= 26; ++i) {
    for (int j = pre[i]; j; j = pre[j])
      if (j == i) {
        puts("NO");
        return 0;
      }
    for (int j = nxt[i]; j; j = nxt[j])
      if (j == i) {
        puts("NO");
        return 0;
      }
  }
  for (int i = 1; i <= 26; ++i)
    if (!pre[i] && vis[i])
      for (int j = i; j; j = nxt[j]) ans.push_back(j);
  for (int i = 0; i < ans.size(); ++i) putchar(ans[i] + 'a' - 1);
  putchar('\n');
  return 0;
}
