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 = 500 + 10;
int n;
string s;
int g[N][N];
int vis[30];
int in[N], out[N];
string ans;
int p[N];
int dfs(int now) {
  vis[now] = 1;
  ans = ans + (char)(now + 'a');
  for (int j = 0; j < 26; ++j) {
    if (g[now][j]) {
      if (vis[j]) return 0;
      return dfs(j);
    }
  }
}
int main() {
  cin >> n;
  int fg = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> s;
    int sz = s.length();
    memset(vis, 0, sizeof(vis));
    for (int j = 0; s[j]; ++j) {
      p[s[j] - 'a'] = 1;
      if (vis[s[j] - 'a'])
        fg = 1;
      else
        vis[s[j] - 'a'] = 1;
      if (j + 1 < s.length()) {
        int now = s[j] - 'a';
        int nxt = s[j + 1] - 'a';
        if (!g[now][nxt]) {
          in[nxt]++;
          out[now]++;
        }
        g[now][nxt] = 1;
      }
    }
  }
  for (int i = 0; i < 26; ++i) {
    fg |= (in[i] > 1 || out[i] > 1);
  }
  if (fg) {
    puts("NO");
    return 0;
  }
  memset(vis, 0, sizeof(vis));
  queue<int> q;
  for (int i = 0; i < 26; ++i) {
    if (out[i] != 0 && (in[i] == 0)) {
      if (dfs(i) == 0) {
        puts("NO");
        return 0;
      }
    } else if (p[i] && !out[i] && !in[i]) {
      ans = ans + (char)('a' + i);
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (in[i] != 0 && out[i] != 0 && !vis[i]) {
      puts("NO");
      return 0;
    }
  }
  for (auto c : ans) {
    cout << c;
  }
}
