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 gra[27][27] = {0}, in[27] = {0}, out[27] = {0};
int main() {
  int n, i, j, f = 0;
  char m[100001], nn[100001];
  scanf("%d", &n);
  int vis[27] = {0};
  for (i = 0; i < n; i++) {
    scanf("%s", m);
    int len = strlen(m);
    vis[m[0] - 'a']++;
    for (j = 0; j < len - 1; j++) {
      int x = m[j] - 'a';
      int y = m[j + 1] - 'a';
      if (!gra[x][y] && (out[x] || in[y])) {
        f = 1;
      }
      if (!gra[x][y]) {
        out[x]++;
        in[y]++;
        gra[x][y] = 1;
        vis[y]++;
      }
    }
  }
  if (f)
    puts("NO");
  else {
    int sum = 0;
    for (i = 0; i < 26 && f; i++) {
      for (j = 0; j < 26 && f; j++) {
        if (gra[i][j] && gra[j][i]) f = 1;
        if (in[i] > 1 || out[i] > 1 || in[j] > 1 || out[j] > 1) f = 1;
      }
    }
    if (f)
      puts("NO");
    else {
      int r = 0, ff = 0, k = 0;
      for (i = 0; i < 26; i++) {
        if (vis[i]) sum++;
        if (vis[i] && !in[i]) {
          nn[k++] = i + 'a';
          int j = 0, ii = i;
          ff = 1;
          while (ii < 26) {
            for (j = 0; j < 26; j++) {
              if (gra[ii][j]) {
                nn[k++] = j + 'a';
                break;
              }
            }
            ii = j;
          }
        }
      }
      nn[k++] = '\0';
      if (!ff || sum != k - 1)
        printf("NO\n");
      else
        puts(nn);
    }
  }
}
