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) % 1000000007LL;
    val = (val * val) % 1000000007LL;
    ex = ex >> 1LL;
  }
  return ans;
}
const int maxn = 1e5 + 10;
vector<int> adj[27], rev[27];
bool visit[27], cycle, reckstack[27], mark[27];
int n, m;
string temp, ans, s1;
set<string> res;
stack<int> s;
void dfs(int start) {
  visit[start] = true;
  reckstack[start] = true;
  for (int i = 0; i < adj[start].size(); i++) {
    int pt = adj[start][i];
    if (!visit[pt]) {
      dfs(pt);
    }
    if (reckstack[pt]) {
      cycle = true;
    }
  }
  reckstack[start] = false;
  s.push(start);
}
void dfs2(int start) {
  visit[start] = true;
  temp += (char)('a' + start);
  for (int i = 0; i < adj[start].size(); i++) {
    int pt = adj[start][i];
    if (!visit[pt]) dfs2(pt);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> s1;
    if (s1.length() == 1) mark[s1[0] - 'a'] = true;
    for (int j = 0; j < s1.length() - 1; j++) {
      int p = s1[j] - 'a';
      int q = s1[j + 1] - 'a';
      if (!adj[p].size())
        adj[p].push_back(q);
      else if (adj[p].size() && adj[p][0] != q)
        cycle = true;
      if (!rev[q].size())
        rev[q].push_back(p);
      else if (rev[q].size() && rev[q][0] != p) {
        cycle = true;
      }
    }
  }
  if (cycle) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < 26; i++) {
    if (!visit[i] && (adj[i].size() || mark[i])) {
      dfs(i);
    }
  }
  if (cycle) {
    cout << "NO" << endl;
    return 0;
  }
  memset(visit, 0, sizeof(visit));
  while (!s.empty()) {
    int p = s.top();
    s.pop();
    if (!visit[p]) {
      temp = "";
      dfs2(p);
      res.insert(temp);
    }
  }
  set<string>::iterator it;
  for (it = res.begin(); it != res.end(); it++) {
    ans += *(it);
  }
  cout << ans << endl;
  return 0;
}
