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;
vector<int> adj[26];
vector<int> vis(26, 0);
int ii = 0;
vector<int> cc(26, 0);
void isc(int i) {
  vis[i] = 1;
  cc[i] = 1;
  for (int j = 0; j < adj[i].size(); j++) {
    if (vis[adj[i][j]] == 0)
      isc(adj[i][j]);
    else if (cc[adj[i][j]] == 1)
      ii = 1;
  }
  cc[i] = 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  vector<string> v(n);
  int u = 0;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    int f[26] = {};
    for (int j = 0; j < v[i].length(); j++) {
      f[v[i][j] - 'a']++;
      if (f[v[i][j] - 'a'] > 1) {
        u = 1;
        break;
      }
    }
  }
  if (u) {
    cout << "NO";
    return 0;
  }
  int cnt[26][26] = {};
  int out[26] = {};
  int in[26] = {};
  int mrk[26] = {};
  for (int i = 0; i < n; i++) {
    mrk[v[i][0] - 'a'] = 1;
    for (int j = 1; j < v[i].size(); j++) {
      mrk[v[i][j] - 'a'] = 1;
      int x = v[i][j - 1] - 'a', y = v[i][j] - 'a';
      if (cnt[x][y] == 0) {
        out[x]++;
        in[y]++;
        adj[x].push_back(y);
      }
      cnt[x][y]++;
    }
  }
  vector<int> st;
  for (int i = 0; i < 26; i++) {
    if (in[i] == 0 && mrk[i] == 1) {
      st.push_back(i);
    }
    if (in[i] > 1 || out[i] > 1) {
      u = 1;
      break;
    }
  }
  if (u) {
    cout << "NO";
    return 0;
  }
  for (int i = 0; i < 26; i++) {
    if (vis[i] == 0) {
      isc(i);
      if (ii) u = 1;
      if (u) break;
    }
  }
  if (u) {
    cout << "NO";
    return 0;
  }
  vector<string> ans;
  for (int i = 0; i < st.size(); i++) {
    int x = st[i];
    string z;
    while (1) {
      z += char('a' + x);
      if (adj[x].empty()) break;
      x = adj[x][0];
    }
    ans.push_back(z);
  }
  sort(ans.begin(), ans.end());
  for (int i = 0; i < ans.size(); i++) cout << ans[i];
  cout << endl;
  return 0;
}
