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 ll = long long;
using ld = long double;
using namespace std;
const int MAXN = 200001;
bool w[26];
int nxt[26];
int pvc[26];
void nein() {
  cout << "NO\n";
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(20);
  cout.tie(nullptr);
  cin.tie(nullptr);
  int n;
  cin >> n;
  fill(begin(nxt), end(nxt), -1);
  for (int i = 0; i < n; ++i) {
    string s;
    cin >> s;
    for (char c : s) {
      w[c - 'a'] = true;
    }
    for (int j = 1; j < s.length(); ++j) {
      int pv = s[j - 1] - 'a', nx = s[j] - 'a';
      if (nxt[pv] == nx) {
        continue;
      }
      ++pvc[nx];
      if (nxt[pv] == -1) {
        nxt[pv] = nx;
        continue;
      }
      nxt[pv] = nx;
    }
  }
  int forcenext = -1;
  string ans;
  for (;;) {
    if (forcenext != -1 && (!w[forcenext] || pvc[forcenext])) {
      nein();
    }
    bool hw = false;
    for (int j = 0; j < 26; ++j) {
      hw |= w[j];
    }
    if (!hw) {
      break;
    }
    int bj;
    if (forcenext != -1) {
      bj = forcenext;
    } else {
      bj = -1;
      for (int j = 25; j >= 0; --j) {
        if (pvc[j] == 0 && w[j]) {
          bj = j;
        }
      }
    }
    if (bj == -1) {
      nein();
    }
    ans += char('a' + bj);
    w[bj] = false;
    forcenext = nxt[bj];
    if (forcenext != -1) {
      --pvc[forcenext];
    }
  }
  cout << ans << "\n";
}
