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 maxn = 100100;
const int maxl = 30;
vector<int> g[maxl], rg[maxl];
string code;
int n, state[maxl];
bool cycle;
bool gc[maxl][maxl], rc[maxl][maxl];
bool visited[maxl];
bool exist[maxl];
void dfs(int node) {
  if (state[node] == 0) {
    state[node] = 1;
    for (int i : g[node]) {
      dfs(i);
    }
    state[node] = 2;
  } else if (state[node] == 1)
    cycle = true;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> code;
    exist[int(code[0] - 'a')] = true;
    for (int j = 1; j < code.length(); j++) {
      if (!gc[int(code[j - 1] - 'a')][int(code[j] - 'a')]) {
        gc[int(code[j - 1] - 'a')][int(code[j] - 'a')] = true;
        g[int(code[j - 1] - 'a')].push_back(int(code[j] - 'a'));
      }
      if (!rc[int(code[j] - 'a')][int(code[j - 1] - 'a')]) {
        rc[int(code[j] - 'a')][int(code[j - 1] - 'a')] = true;
        rg[int(code[j] - 'a')].push_back(int(code[j - 1] - 'a'));
      }
      exist[int(code[j] - 'a')] = true;
    }
  }
  for (int i = 0; i < 30; i++) {
    if (g[i].size() > 1 || rg[i].size() > 1) {
      cout << "NO\n";
      return 0;
    }
  }
  for (int i = 0; i < 26; i++) {
    dfs(i);
  }
  if (cycle) {
    cout << "NO\n";
    return 0;
  }
  vector<string> paths;
  for (int i = 0; i < 26; i++) {
    if (!visited[i] && rg[i].size() == 0 && exist[i]) {
      int x = i;
      string curr = "";
      while (g[x].size() > 0) {
        curr += char(x + 'a');
        x = g[x][0];
      }
      curr += char(x + 'a');
      paths.push_back(curr);
    }
  }
  sort(paths.begin(), paths.end());
  for (auto i : paths) {
    cout << i;
  }
  cout << "\n";
  return 0;
}
