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 = 1100;
int n;
char a[110000];
int vis[26];
int nex[26], up[26];
int main() {
  memset(nex, -1, sizeof(nex));
  memset(up, -1, sizeof(up));
  scanf("%d", &n);
  int flag = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%s", a);
    int len = strlen(a);
    for (int j = 0; j < len - 1; j++) {
      vis[a[j] - 'a'] = 1;
      if (nex[a[j] - 'a'] == -1 && up[a[j + 1] - 'a'] == -1) {
        nex[a[j] - 'a'] = a[j + 1] - 'a';
        up[a[j + 1] - 'a'] = a[j] - 'a';
      } else {
        if (nex[a[j] - 'a'] != a[j + 1] - 'a' ||
            up[a[j + 1] - 'a'] != a[j] - 'a') {
          flag = 1;
        }
      }
    }
    vis[a[len - 1] - 'a'] = 1;
  }
  int cnt = 0;
  for (int i = 0; i <= 25; i++) {
    if (vis[i]) {
      cnt++;
    }
  }
  if (flag)
    printf("NO\n");
  else {
    char ans[27];
    int cn = 0;
    priority_queue<int, vector<int>, greater<int> > q;
    for (int i = 0; i <= 25; i++) {
      if ((up[i] == -1 && nex[i] != -1)) q.push(i);
    }
    for (int i = 0; i <= 25; i++) {
      if (up[i] == -1 && nex[i] == -1 && vis[i]) q.push(i);
    }
    while (!q.empty()) {
      int t = q.top();
      q.pop();
      ans[cn++] = t + 'a';
      for (int i = i = nex[t]; i != -1; i = nex[i]) ans[cn++] = i + 'a';
    }
    ans[cnt] = '\0';
    if (cnt == cn)
      cout << ans << endl;
    else
      printf("NO\n");
  }
  return 0;
}
