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 inf = 1000000007;
const long long linf = 1ll * inf * inf;
const int N = 100000 + 7;
const int M = 30;
const int multipleTest = 0;
int n;
string s[N];
int last[26];
int nxt[100];
bool valid[100];
bool has[100];
int cnt[100];
bool check[100];
void solve() {
  cin >> n;
  for (int j = (0), _b = (26); j < _b; ++j) nxt[j] = -1;
  for (int i = (0), _b = (n); i < _b; ++i) {
    cin >> s[i];
    for (int j = (0), _b = (26); j < _b; ++j) last[j] = -1;
    for (char c : s[i]) {
      has[c - 'a'] = true;
      if (last[c - 'a'] != -1) {
        puts("NO");
        return;
      }
      last[c - 'a'] = 0;
    }
    for (int j = (0), _b = ((int)(s[i]).size() - 1); j < _b; ++j) {
      int c = s[i][j] - 'a';
      int u = s[i][j + 1] - 'a';
      if (nxt[c] == -1 || nxt[c] == u) {
        nxt[c] = u;
        valid[u] = true;
      } else {
        puts("NO");
        return;
      }
    }
  }
  for (int j = (0), _b = (26); j < _b; ++j)
    if (has[j] && nxt[j] != -1) cnt[nxt[j]]++;
  for (int j = (0), _b = (26); j < _b; ++j)
    if (cnt[j] > 1) {
      puts("NO");
      return;
    }
  for (int j = (0), _b = (26); j < _b; ++j)
    if (has[j] && cnt[j] == 1) {
      int tmp = nxt[j];
      while (tmp != -1) {
        if (tmp == j) {
          puts("NO");
          return;
        }
        tmp = nxt[tmp];
      }
    }
  for (int j = (0), _b = (26); j < _b; ++j)
    if (has[j] && !valid[j]) {
      int tmp = j;
      while (tmp != -1) {
        printf("%c", char(tmp + 'a'));
        valid[tmp] = true;
        tmp = nxt[tmp];
      }
    }
}
int main() {
  int Test = 1;
  if (multipleTest) {
    cin >> Test;
  }
  for (int i = 0; i < Test; ++i) {
    solve();
  }
}
