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>
#pragma comment(linker, "/stack:16777216")
using namespace std;
const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const long long LINF = INF * (long long)INF;
int n;
string s[200000];
vector<pair<int, int> > q[30];
int M[30][30];
bool used[30];
bool used2[30];
int main() {
  cin >> n;
  for (int i = (0); i < (n); i++) {
    cin >> s[i];
    if (s[i].size() > 26) {
      cout << "NO";
      return 0;
    }
    for (int j = (0); j < (s[i].size()); j++) used2[s[i][j] - 'a'] = 1;
    for (int j = (1); j < (s[i].size()); j++) {
      M[s[i][j - 1] - 'a'][s[i][j] - 'a'] = 1;
    }
  }
  for (int i = (0); i < (30); i++) {
    int q1 = 0, q2 = 0;
    for (int j = (0); j < (30); j++) {
      if (M[i][j] && M[j][i]) {
        cout << "NO";
        return 0;
      }
      if (M[j][i]) q1++;
      if (M[i][j]) q2++;
    }
    if (q1 > 1 || q2 > 1) {
      cout << "NO";
      return 0;
    }
  }
  int Y = 0;
  for (int i = (0); i < (30); i++) Y += used2[i];
  string ans = "";
  for (int st = (0); st < (30); st++)
    if (!used[st] && used2[st]) {
      bool ok = 1;
      for (int e = (0); e < (30); e++)
        if (M[e][st]) ok = 0;
      if (!ok) continue;
      int pos = st;
      do {
        ans += char('a' + pos);
        used[pos] = 1;
        int nxt = -1;
        for (int z = (0); z < (30); z++) {
          if (!used[z] && M[pos][z]) nxt = z;
          if (used[z] && M[pos][z]) {
            cout << "NO";
            return 0;
          }
        }
        pos = nxt;
      } while (pos != -1);
    }
  if (ans.size() != Y) ans = "NO";
  cout << ans;
}
