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;
bool lines = 1, vis[26] = {}, isstring[26] = {}, let[26] = {};
int g[26][26] = {};
string ans;
void dfs(int u) {
  vis[u] = 1;
  ans += (char)(u + 'a');
  int cnt = 0;
  for (int j = 0; j < 26; j++) {
    if (!g[u][j]) {
      continue;
    }
    cnt++;
    if (!vis[j]) {
      dfs(j);
    } else {
      lines = 0;
    }
  }
  if (cnt > 1) {
    lines = 0;
  }
}
int main() {
  int n;
  cin >> n;
  vector<string> a(n);
  for (auto &x : a) {
    cin >> x;
    for (int j = 0; j < x.size() - 1; j++) {
      g[x[j] - 'a'][x[j + 1] - 'a'] = 1;
      let[x[j] - 'a'] = 1;
      let[x[j + 1] - 'a'] = 1;
    }
    if (x.size() == 1) {
      isstring[x[0] - 'a'] = 1;
    }
  }
  for (int i = 0; i < 26; i++) {
    int cntin = 0, cntout = 0;
    for (int j = 0; j < 26; j++) {
      cntin += g[j][i];
      cntout += g[i][j];
    }
    if (cntin + cntout == 0 && !isstring[i]) {
      continue;
    }
    if (cntin == 0 && !vis[i]) {
      dfs(i);
    }
  }
  for (int i = 0; i < 26; i++) {
    if (let[i]) {
      if (count(ans.begin(), ans.end(), (char)('a' + i)) == 0) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  if (ans.empty() || !lines) {
    cout << "NO" << endl;
  } else {
    cout << ans << endl;
  }
  return 0;
}
