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 nex[30], pre[30];
char ans[30];
const int maxn = 1e5 + 5;
int main() {
  int n, tol = 0;
  char a[maxn];
  bool can = true;
  memset(nex, -1, sizeof nex);
  memset(pre, -1, sizeof pre);
  scanf("%d", &n);
  getchar();
  for (int i = 0; i < n; i++) {
    scanf("%s", a);
    int lena = strlen(a);
    if (pre[a[0] - 'a'] == -1) pre[a[0] - 'a'] = -2;
    for (int i = 1; i < lena; i++) {
      if ((nex[a[i - 1] - 'a'] == -1 || nex[a[i - 1] - 'a'] == a[i] - 'a') &&
          (pre[a[i] - 'a'] == -1 || pre[a[i] - 'a'] == -2 ||
           pre[a[i] - 'a'] == a[i - 1] - 'a')) {
        nex[a[i - 1] - 'a'] = a[i] - 'a';
        pre[a[i] - 'a'] = a[i - 1] - 'a';
      } else {
        can = false;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (pre[i] == -2) {
      for (int k = i; k != -1;) {
        ans[tol++] = k + 'a';
        k = nex[k];
      }
    }
  }
  a[tol++] = '\0';
  for (int i = 0; i < 26; i++) {
    for (int k = nex[i]; k != -1;) {
      if (k == i) {
        can = false;
        break;
      }
      k = nex[k];
    }
  }
  if (!can)
    printf("NO\n");
  else
    printf("%s\n", ans);
}
