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 = 100005;
int n, Ans, in[N], out[N], g[26][26], tag[26], vis[26], inq[26];
char s[N], ans[30];
int dfs1(int x) {
  ans[++Ans] = 'a' + x;
  for (int i = 0; i < 26; ++i)
    if (g[x][i]) dfs1(i);
}
int dfs(int x) {
  vis[x] = inq[x] = 1;
  for (int i = 0; i < 26; ++i)
    if (g[x][i]) {
      if (inq[i] || !vis[i] && dfs(i)) return 1;
    }
  return inq[x] = 0;
}
int main() {
  int i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%s", s);
    for (j = 1; s[j]; ++j) {
      int x = s[j - 1] - 'a', y = s[j] - 'a';
      if (g[x][y]) continue;
      g[x][y] = 1;
      ++out[x];
      ++in[y];
    }
    if (j == 1) tag[s[0] - 'a']++;
  }
  for (i = 0; i < 26; ++i)
    if (in[i] > 1 || out[i] > 1 || g[i][i]) return puts("NO"), 0;
  for (i = 0; i < 26; ++i) {
    if (in[i] + tag[i] > 0 && !vis[i] && dfs(i)) return puts("NO"), 0;
  }
  for (i = 0; i < 26; ++i) {
    if (!in[i] && out[i])
      dfs1(i);
    else if (!in[i] && !out[i] && tag[i])
      ans[++Ans] = 'a' + i;
  }
  puts(ans + 1);
  return 0;
}
