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;
string vow = "aeiou";
int month[] = {-1, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int dxhorse[] = {-2, -2, -1, -1, 1, 1, 2, 2};
const int dyhorse[] = {1, -1, 2, -2, 2, -2, 1, -1};
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
const long double pie = 3.1415926535897932384626;
const long long mod = 1e9 + 7;
int n;
const int N = 26;
vector<int> g[N];
int in_deg[N], out_deg[N];
bool present[N];
void bad() {
  cout << "NO";
  cout << "\n";
  ;
  exit(0);
}
void find_edges(string &s) {
  int z = s.size();
  for (auto i : s) present[i - 'a'] = 1;
  for (int j = 1; j < z; j++) {
    int i = j - 1;
    if (s[i] == s[j]) bad();
    int a = s[i] - 'a';
    int b = s[j] - 'a';
    if (g[a].size() > 0 && g[a].back() == b) continue;
    ;
    g[a].push_back(b);
    out_deg[a]++;
    in_deg[b]++;
  }
}
void read() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    int z = s.size();
    if (z > 26) bad();
    find_edges(s);
  }
}
bool vis[N];
bool in_stack[N];
void dfs_cycle(int node) {
  vis[node] = 1;
  in_stack[node] = 1;
  for (auto i : g[node]) {
    if (in_stack[i]) bad();
    if (vis[i] == false) dfs_cycle(i);
  }
  in_stack[node] = false;
}
void check_if_cycle() {
  for (int i = 0; i < N; i++) {
    if (vis[i]) continue;
    ;
    dfs_cycle(i);
  }
}
void check_degree() {
  for (int i = 0; i < N; i++) {
    if (in_deg[i] > 1) bad();
    if (out_deg[i] > 1) bad();
  }
}
void FILL() {
  for (int i = 0; i < N; i++) vis[i] = false;
}
void dfs(int node) {
  vis[node] = 1;
  char c = 'a' + node;
  cout << c;
  for (auto i : g[node]) {
    if (vis[i]) continue;
    ;
    dfs(i);
  }
}
void print() {
  FILL();
  for (int i = 0; i < N; i++) {
    if (present[i] == false) continue;
    ;
    if (present[i] == 1 && (in_deg[i] + out_deg[i] == 0))
      dfs(i);
    else if (out_deg[i] == 1 && in_deg[i] == 0) {
      dfs(i);
    }
  }
}
void solve(int test_case) {
  read();
  check_if_cycle();
  check_degree();
  print();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  for (int i = 1; i <= t; i++) solve(i);
  return 0;
}
