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[27];
vector<int> rev[27];
bool mark[27], cycle;
vector<int> stk;
int visited[27];
int N;
vector<string> strs;
string tmp;
void dfs(int u) {
  visited[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (visited[v] == 0) {
      dfs(v);
    } else if (visited[v] == 1) {
      cycle = true;
    }
  }
  visited[u] = 2;
  stk.push_back(u);
}
void dfs2(int u) {
  visited[u] = 1;
  tmp += (char)('a' + u);
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (visited[v] == 0) {
      dfs2(v);
    }
  }
  visited[u] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> N;
  cycle = false;
  string s;
  for (int i = 0; i < 27; i++) {
    mark[i] = false;
    visited[i] = 0;
  }
  for (int i = 0; i < N; i++) {
    cin >> s;
    strs.push_back(s);
    if (s.length() == 1) {
      mark[s[0] - 'a'] = true;
    }
    for (int j = 0; j < s.length() - 1; j++) {
      int p = s[j] - 'a';
      int q = s[j + 1] - 'a';
      if (!adj[p].size()) {
        adj[p].push_back(q);
      } else if (adj[p].size() && adj[p][0] != q) {
        cycle = true;
      }
      if (!rev[q].size()) {
        rev[q].push_back(p);
      } else if (rev[q].size() && rev[q][0] != p) {
        cycle = true;
      }
      if (cycle) {
        cout << "NO" << endl;
        return 0;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (!visited[i] && (adj[i].size() || mark[i])) {
      dfs(i);
    }
  }
  if (cycle) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < 26; i++) visited[i] = 0;
  reverse(stk.begin(), stk.end());
  set<string> res;
  for (int i = 0; i < stk.size(); i++) {
    if (visited[stk[i]] == 0) {
      tmp = "";
      dfs2(stk[i]);
      res.insert(tmp);
    }
  }
  set<string>::iterator it;
  string ans = "";
  for (it = res.begin(); it != res.end(); it++) {
    ans += *it;
  }
  cout << ans << endl;
}
