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 long long sz = 3e5 + 5, mod = 1e9 + 7;
long long t, n, m, k, q;
long long x, y, in, fn;
long long mx = -1, L;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  t = 1;
  while (t--) {
    cin >> n;
    vector<long long> vis(27, 0), pre(27, -1), nxt(27, -1);
    bool ok = 1;
    for (long long i = 1; i <= n; i++) {
      string s;
      cin >> s;
      for (long long j = 1; s[j]; j++) {
        long long x = s[j - 1] - 'a';
        long long y = s[j] - 'a';
        if ((nxt[x] != -1 && nxt[x] != y) || (pre[y] != -1 && pre[y] != x))
          ok = 0;
        pre[y] = x;
        nxt[x] = y;
      }
      for (long long j = 0; s[j]; j++) vis[s[j] - 'a'] = 1;
    }
    if (!ok) {
      cout << "NO";
      return 0;
    }
    long long s = 0;
    for (long long i = 0; i < 26; i++)
      if (vis[i]) s++;
    for (long long i = 0; i < 26; i++) {
      if (vis[i] && pre[i] == -1) {
        for (long long j = i; j != -1; j = nxt[j]) {
          s--;
        }
      }
    }
    if (s != 0) {
      cout << "NO";
      return 0;
    }
    for (long long i = 0; i < 26; i++) {
      if (vis[i] && pre[i] == -1) {
        for (long long j = i; j != -1; j = nxt[j]) {
          char c = 'a' + j;
          cout << c;
        }
      }
    }
  }
  return 0;
}
