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 LOG = 18;
const int NN = 2e5 + 5;
const int MOD = (int)1e9 + 7;
const int INF = 1LL << 28;
bool adj[30][30];
char clr[30];
int in_deg[30];
bool exist[30];
string ans;
bool good;
void dfs(int x) {
  if (clr[x] == 'g') {
    good = false;
    return;
  }
  if (clr[x] != 'w') return;
  clr[x] = 'g';
  ans += ((char)(x + 'a'));
  for (int y = 0; y < 30; y++)
    if (adj[x][y]) {
      dfs(y);
    }
  clr[x] = 'b';
}
char s[1 << 17];
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  scanf("%d\n", &n);
  memset(adj, false, sizeof(adj));
  while (n--) {
    fgets(s, 1 << 19, stdin);
    int l = strlen(s);
    if (s[l - 1] == '\n') s[l - 1] = '\0', l--;
    for (int i = 0; i < l; i++) {
      exist[s[i] - 'a'] = true;
      if (i > 0) adj[s[i - 1] - 'a'][s[i] - 'a'] = true;
    }
  }
  good = true;
  ans.clear();
  for (int i = 0; i < 30; i++) {
    in_deg[i] = 0;
    int out = 0;
    for (int j = 0; j < 30; j++) {
      in_deg[i] += adj[j][i];
      out += adj[i][j];
    }
    if (in_deg[i] >= 2 || out >= 2) {
      good = false;
    }
  }
  memset(clr, 'w', sizeof(clr));
  for (int i = 0; i < 30; i++)
    if (exist[i] == true && in_deg[i] == 0) {
      dfs(i);
    }
  for (int i = 0; i < 30; i++)
    if (exist[i] == true && clr[i] == 'w') good = false;
  if (good)
    printf("%s\n", ans.c_str());
  else
    printf("NO\n");
  return 0;
}
