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 maxn = 100000 + 5;
string s[maxn];
int nxt[30];
int pre[30];
bool vis[30];
bool exist[30];
void dfs(int x) {
  vis[x] = true;
  cout << (char)('a' + x - 1);
  if (nxt[x] == 0) {
    return;
  }
  dfs(nxt[x]);
}
bool flag = true;
void dfs2(int x) {
  if (vis[x] || flag == false) {
    flag = false;
    return;
  }
  vis[x] = true;
  if (nxt[x] == 0) {
    return;
  }
  dfs2(nxt[x]);
}
int main(void) {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> s[i];
    for (size_t j = 0; j < s[i].length(); j++) {
      exist[s[i][j] - 'a' + 1] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    for (size_t j = 0; j < s[i].length() - 1; j++) {
      if (nxt[s[i][j] - 'a' + 1] == 0 && pre[s[i][j + 1] - 'a' + 1] == 0) {
        nxt[s[i][j] - 'a' + 1] = s[i][j + 1] - 'a' + 1;
        pre[s[i][j + 1] - 'a' + 1] = s[i][j] - 'a' + 1;
      } else if (nxt[s[i][j] - 'a' + 1] == s[i][j + 1] - 'a' + 1 &&
                 pre[s[i][j + 1] - 'a' + 1] == s[i][j] - 'a' + 1) {
        continue;
      } else {
        cout << "NO";
        return 0;
      }
    }
  }
  int cnt = 0;
  for (int i = 1; i <= 26; i++) {
    if (exist[i] && !vis[i] && pre[i] == 0) {
      cnt++;
      dfs2(i);
    }
  }
  for (int i = 1; i <= 26; i++) {
    if (exist[i] && !vis[i]) {
      flag = false;
      break;
    }
  }
  if (flag == false || cnt == 0) {
    cout << "NO";
    return 0;
  }
  memset(vis, false, sizeof vis);
  for (int i = 1; i <= 26; i++) {
    if (exist[i] && !vis[i] && pre[i] == 0) {
      dfs(i);
    }
  }
  return 0;
}
