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 = 1.1e5;
const int MOD = (int)(1e9) + 7;
using ll = long long;
vector<int> G[26];
bool stk[26];
int link[26];
bool vis[26];
string ans;
bool dfs(int n) {
  if (stk[n]) return false;
  if (vis[n]) return true;
  stk[n] = true;
  vis[n] = true;
  bool ok = G[n].empty() || dfs(G[n][0]);
  ans = (char)('a' + n) + ans;
  stk[n] = false;
  return ok;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  memset(link, -1, sizeof(link));
  vector<int> indeg(100);
  vector<int> outdeg(100);
  set<int, greater<int> > sources;
  vector<int> toolazy;
  for (int i = 0; i < n; i++) {
    string X;
    cin >> X;
    sources.insert(X[0] - 'a');
    if (X.size() > 1) toolazy.push_back(X[0] - 'a');
    for (int j = 0; j + 1 < X.size(); j++) {
      int u = X[j] - 'a';
      int v = X[j + 1] - 'a';
      if (link[u] == v || link[u] == -1) {
        link[u] = v;
      } else {
        cout << "NO\n";
        exit(0);
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (link[i] != -1) {
      indeg[link[i]]++;
      outdeg[i]++;
      G[i].push_back(link[i]);
    }
  }
  for (int x : indeg) {
    if (x >= 2) {
      cout << "NO\n";
      exit(0);
    }
  }
  for (int x : outdeg) {
    if (x >= 2) {
      cout << "NO\n";
      exit(0);
    }
  }
  vector<string> paths;
  sort(toolazy.begin(), toolazy.end(),
       [&](int x, int y) { return indeg[x] < indeg[y]; });
  for (int i : toolazy) {
    if (G[i].size()) ans.clear();
    if (G[i].size() && !dfs(i)) {
      cout << "NO\n";
      exit(0);
    }
    paths.push_back(ans);
  }
  for (int i : sources) {
    if (!vis[i]) {
      string s;
      s += (char)(i + 'a');
      paths.push_back(s);
    }
  }
  sort(paths.begin(), paths.end());
  for (string s : paths) cout << s;
  cout << '\n';
  return 0;
}
