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 N = 1e5 + 5, INF = 1e9 + 7, md = INF;
string s[N];
int freq[28], par[N];
vector<int> idx[N];
void init(int n) {
  for (int i = 1; i <= n; i++) {
    par[i] = i;
  }
}
int findPar(int u) {
  if (par[u] == u) return u;
  return par[u] = findPar(par[u]);
}
void Merge(int u, int v, int ch) {
  u = findPar(u), v = findPar(v);
  if (u == v) return;
  int idx1 = -1, idx2 = -1;
  for (int i = 0; i < s[u].size(); i++) {
    if (s[u][i] == ch + 'a') {
      idx1 = i;
      break;
    }
  }
  for (int i = 0; i < s[v].size(); i++) {
    if (s[v][i] == ch + 'a') {
      idx2 = i;
      break;
    }
  }
  for (int i = idx1, j = idx2; i < s[u].size() and j < s[v].size(); i++, j++) {
    if (s[u][i] != s[v][j]) {
      cout << "NO";
      exit(0);
    }
  }
  for (int i = idx1 - 1, j = idx2 - 1; i >= 0 and j >= 0; i--, j--) {
    if (s[u][i] != s[v][j]) {
      cout << "NO";
      exit(0);
    }
  }
  if (idx1 > idx2) {
    swap(idx1, idx2);
    swap(u, v);
  }
  par[u] = v;
  int sz = max((int)s[v].size(), idx2 + (int)s[u].size() - idx1);
  if (sz > 26) {
    cout << "NO";
    exit(0);
  }
  int second = (int)s[v].size();
  s[v].resize(sz);
  for (int i = sz - 1, j = s[u].size() - 1; i >= second; i--, j--) {
    s[v][i] = s[u][j];
  }
}
int main() {
  int n;
  cin >> n;
  init(n);
  for (int i = 1; i <= n; i++) {
    cin >> s[i];
    memset(freq, 0, sizeof freq);
    for (int j = 0; j < s[i].size(); j++) {
      freq[s[i][j] - 'a']++;
    }
    for (int j = 0; j < 26; j++) {
      if (freq[j] > 1) {
        cout << "NO";
        return 0;
      }
      if (freq[j]) {
        idx[j].push_back(i);
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    for (int j = 1; j < idx[i].size(); j++) {
      int u = idx[i][j - 1], v = idx[i][j];
      Merge(u, v, i);
    }
  }
  vector<string> res;
  int flag = 1;
  memset(freq, 0, sizeof freq);
  for (int i = 1; i <= n; i++) {
    if (par[i] == i) {
      res.push_back(s[i]);
      for (int j = 0; j < s[i].size(); j++) {
        freq[s[i][j] - 'a']++;
      }
    }
  }
  for (int i = 0; i < 26; i++) {
    if (freq[i] > 1) {
      flag = 0;
    }
  }
  if (flag == 0) {
    cout << "NO";
    exit(0);
  }
  sort(res.begin(), res.end());
  for (int i = 0; i < res.size(); i++) {
    cout << res[i];
  }
}
