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 n;
string a[100007];
int nxt[26];
bool was[26];
bool can[26];
bool need[26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    set<char> se;
    for (char c : a[i]) {
      if (se.find(c) != se.end()) cout << "NO", exit(0);
      se.insert(c);
      need[c - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++) nxt[i] = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < (int)a[i].size() - 1; j++) {
      int c1 = a[i][j] - 'a';
      int c2 = a[i][j + 1] - 'a';
      if (nxt[c1] != -1 && nxt[c1] != c2) cout << "NO", exit(0);
      nxt[c1] = c2;
    }
    for (int j = 1; j < a[i].size(); j++) can[a[i][j] - 'a'] = 1;
  }
  for (int i = 0; i < 26; i++)
    for (int j = 0; j < 26; j++)
      if (i != j && nxt[i] != -1 && nxt[j] != -1) {
        if (nxt[i] == nxt[j]) cout << "NO", exit(0);
      }
  string ans;
  for (int i = 0; i < 26; i++)
    if (need[i] && !can[i] && !was[i]) {
      int v = i;
      while (1) {
        if (was[v]) break;
        ans += (char)(v + 'a');
        was[v] = 1;
        v = nxt[v];
        if (v == -1) break;
      }
    }
  for (int i = 0; i < n; i++) {
    bool f = 0;
    for (int j = 0; j < ans.size(); j++) {
      bool ff = 1;
      for (int s = 0; s < a[i].size(); s++) {
        if (j + s >= ans.size()) {
          ff = 0;
          break;
        }
        if (a[i][s] != ans[j + s]) ff = 0;
      }
      if (ff) f = 1;
    }
    if (!f) cout << "NO", exit(0);
  }
  cout << ans;
}
