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;
char str[1000001];
int n;
vector<string> v;
vector<vector<int> > g;
vector<bool> vis;
vector<string> sol;
void DFS(int u) {
  if (vis[u]) {
    puts("NO");
    exit(0);
  }
  vis[u] = true;
  sol.back() += char(u + 'a');
  if (g[u].empty()) return;
  DFS(g[u][0]);
}
int main() {
  scanf("%d", &n);
  vector<bool> seen(26);
  for (int i = 0; i < n; ++i) {
    scanf("%s", str);
    v.push_back(str);
    set<char> s;
    for (int i = 0; str[i]; ++i) {
      s.insert(str[i]);
      seen[str[i] - 'a'] = true;
    }
    if (s.size() != v.back().size()) {
      puts("NO");
      return 0;
    }
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  if (v.size() > 700) {
    puts("NO");
    return 0;
  }
  for (int it = 0; it < 2; ++it) {
    for (int i = 0; i < v.size(); ++i)
      for (int j = i + 1; j < v.size(); ++j) {
        if ((int)v[i].find(v[j]) != -1) {
          v[j] = v.back();
          v.pop_back();
          --j;
        }
      }
    reverse(v.begin(), v.end());
  }
  g.resize(26);
  vector<int> in(g.size());
  set<pair<int, int> > e;
  for (int i = 0; i < v.size(); ++i) {
    string s = v[i];
    for (int j = 0; j + 1 < s.size(); ++j) {
      int u = s[j] - 'a';
      int v = s[j + 1] - 'a';
      if (e.insert(make_pair(u, v)).second) {
        g[u].push_back(v);
        ++in[v];
      }
    }
  }
  for (int i = 0; i < g.size(); ++i) {
    if (g[i].size() > 1) {
      puts("NO");
      return 0;
    }
  }
  vis.resize(g.size());
  for (int i = 0; i < g.size(); ++i)
    if (seen[i] && !vis[i] && in[i] == 0) {
      sol.push_back(string());
      DFS(i);
    }
  for (int i = 0; i < g.size(); ++i)
    if (seen[i] && !vis[i]) {
      puts("NO");
      return 0;
    }
  sort(sol.begin(), sol.end());
  string ans = "";
  for (auto x : sol) ans += x;
  set<char> zz;
  for (auto x : ans) zz.insert(x);
  if (zz.size() != ans.size())
    ans = "NO";
  else {
    set<int> S;
    for (int i = 0; i < v.size(); ++i) {
      int occ = 0;
      for (int j = 0; j + v[i].size() <= ans.size(); ++j)
        if (ans.substr(j, v[i].size()) == v[i]) ++occ;
      if (!occ) {
        ans = "NO";
        break;
      }
      S.insert(occ);
    }
    if (S.size() != 1) ans = "NO";
  }
  cout << ans << endl;
  return 0;
}
