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 = 100100;
char *s[maxn], *p, buf[maxn * 2];
char t[maxn];
int alp[30][30], d[30], is[30], vis[30];
int ct;
bool dfs(int u) {
  t[ct++] = u + 'a';
  vis[u] = 1;
  int cnt = 0, p = -1;
  for (int i = 0; i < 26; i++) {
    if (alp[u][i]) cnt++, p = i;
  }
  if (cnt > 1) return false;
  if (cnt == 0) {
    t[ct] = 0;
    return true;
  }
  if (vis[p]) return 0;
  return dfs(p);
}
int main(void) {
  int n;
  scanf("%d", &n);
  p = buf;
  for (int i = 1; i <= n; i++) {
    scanf("%s", p);
    s[i] = p;
    p += strlen(p) + 1;
  }
  for (int i = 1; i <= n; i++) {
    is[s[i][0] - 'a'] = 1;
    for (int j = 1; s[i][j]; j++) {
      alp[s[i][j - 1] - 'a'][s[i][j] - 'a'] = 1;
      d[s[i][j] - 'a']++;
      is[s[i][0] - 'a'] = 1;
    }
  }
  int flag = 0, yes = 0;
  for (int i = 0; i < 26; i++)
    if (is[i] == 1) {
      if (d[i] == 0) {
        yes = 1;
        if (dfs(i) == false) {
          flag = 1;
          break;
        }
      }
    }
  for (int i = 0; i < 26; i++)
    if (is[i] && vis[i] == 0) {
      flag = 1;
    }
  if (flag || yes == 0) {
    printf("NO\n");
    return 0;
  }
  printf("%s\n", t);
  return 0;
}
