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;
int gcd1(int a, int b) {
  if (a == 0) return b;
  return gcd1(b % a, a);
}
long long modx(long long base, long long ex) {
  long long ans = 1LL, val = base;
  while (ex > 0LL) {
    if (ex & 1LL) ans = (ans * val) % 1000000009LL;
    val = (val * val) % 1000000009LL;
    ex = ex >> 1LL;
  }
  return ans;
}
int n;
vector<string> v;
set<int> adj[27];
set<int> adj2[27];
bool visit[27];
bool recstack[27];
bool mark[27];
bool cycle;
stack<int> s;
string str;
string ans = "";
void dfs2(int start) {
  ans += (char)(start + 'a');
  visit[start] = true;
  if (!adj[start].size()) return;
  dfs2(*adj[start].begin());
}
void dfs(int start) {
  if (cycle) return;
  visit[start] = true;
  recstack[start] = true;
  if (!adj[start].size()) {
    recstack[start] = false;
    s.push(start);
    return;
  }
  int pt = *adj[start].begin();
  if (recstack[pt]) {
    cycle = true;
    return;
  }
  if (!visit[pt]) dfs(pt);
  recstack[start] = false;
  s.push(start);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> str;
    v.push_back(str);
  }
  for (int i = 0; i < n; i++) {
    str = v[i];
    int l = str.length();
    for (int j = 0; j < l - 1; j++) {
      adj[str[j] - 'a'].insert(str[j + 1] - 'a');
      adj2[str[j + 1] - 'a'].insert(str[j] - 'a');
      if (adj[str[j] - 'a'].size() > 1 || adj2[str[j + 1] - 'a'].size() > 1) {
        cout << "NO" << endl;
        return 0;
      }
      mark[str[j] - 'a'] = true;
    }
    mark[str[l - 1] - 'a'] = true;
  }
  sort(v.rbegin(), v.rend());
  for (int i = 25; i >= 0; i--) {
    if (mark[i] && !visit[i]) dfs(i);
    if (cycle) {
      cout << "NO" << endl;
      return 0;
    }
  }
  memset(visit, false, sizeof(visit));
  while (!s.empty()) {
    int t = s.top();
    s.pop();
    if (visit[t]) continue;
    dfs2(t);
  }
  cout << ans << endl;
  return 0;
}
