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> v[30];
bool vis[30], vis2[30], cnt[30];
int deg[30];
string P;
deque<int> st;
vector<string> ans;
int mp[30][30], before[30];
bool dfs(int u) {
  vis[u] = 1;
  vis2[u] = 1;
  bool fl = 0;
  for (int i : v[u]) {
    if (vis[i]) return 1;
    fl |= dfs(i);
  }
  st.push_front(u);
  vis[u] = 0;
  return fl;
}
void fn(int u) {
  vis[u] = 1;
  P.push_back(u + 'a');
  for (int i : v[u]) {
    if (vis[i]) continue;
    fn(i);
  }
}
int main() {
  memset(before, -1, sizeof before);
  bool fl = 0;
  int TC;
  cin >> TC;
  while (TC--) {
    string s;
    cin >> s;
    for (int j = 1; j < s.size(); j++) {
      int a = s[j - 1] - 'a', b = s[j] - 'a';
      deg[a]++;
      deg[b]++;
      cnt[a] = cnt[b] = 1;
      if (mp[a][b]) continue;
      mp[a][b] = 1;
      if (before[b] != -1 && before[b] != a) fl = 1;
      before[b] = a;
      v[a].push_back(b);
    }
    if (s.size() == 1) cnt[s[0] - 'a'] = 1, deg[s[0] - 'a']++;
  }
  for (int i = 0; i < 26; i++) {
    if (vis2[i] || cnt[i] == false) continue;
    fl |= dfs(i);
  }
  if (fl) {
    puts("NO");
    return 0;
  }
  for (int i = 0; i < 26; i++) {
    int x = 1;
    for (int j = 1; j < v[i].size(); j++)
      if (v[i][j] != v[i][j - 1]) {
        puts("NO");
        return 0;
      }
  }
  while (st.size()) {
    int u = st.front();
    st.pop_front();
    if (vis[u]) continue;
    P = "";
    fn(u);
    int a = 0;
    for (int i = 0; i < P.size(); i++)
      if (deg[P[i] - 'a'] == 1) a++;
    if (a > 2) {
      puts("NO");
      return 0;
    }
    ans.push_back(P);
  }
  sort(ans.begin(), ans.end());
  for (auto i : ans) cout << i;
  return 0;
}
