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;
int n;
char ch[MAXN];
int NEXT[27], PREV[27], vis[27], ans[MAXN], cnt[27], Index, tmp[27];
queue<int> Q;
int init() {
  scanf("%d", &n);
  for (int o = 1; o <= n; o++) {
    scanf(" %s", ch + 1);
    int len = strlen(ch + 1);
    for (int i = 1; i <= len; i++) {
      int u = ch[i] - 'a' + 1, v = ch[i + 1] - 'a' + 1;
      vis[u] = true;
      if (i == len) continue;
      if (NEXT[u] == 0) NEXT[u] = v;
      if (PREV[v] == 0) PREV[v] = u;
      if (NEXT[u] != v) return printf("NO\n"), 0;
      if (PREV[v] != u) return printf("NO\n"), 0;
    }
  }
  while (Q.empty() == false) Q.pop();
  for (int i = 1; i <= 26; i++)
    if (PREV[i] == 0 && vis[i] == true) {
      Q.push(i);
    }
  for (int i = 1; i <= 26; i++) tmp[i] = vis[i];
  for (int i = 1; i <= 26; i++) vis[i] = false;
  if (Q.empty() == true) return printf("NO\n"), 0;
  while (Q.empty() == false) {
    int u = Q.front();
    Q.pop();
    ans[++Index] = u;
    cnt[u]++;
    if (vis[u] == true) return printf("NO\n"), 0;
    vis[u] = true;
    while (NEXT[u] != 0) {
      u = NEXT[u];
      if (vis[u] == true) return printf("NO\n"), 0;
      vis[u] = true;
      ans[++Index] = u;
      cnt[u]++;
    }
  }
  for (int i = 1; i <= 26; i++)
    if (cnt[i] > 1) return printf("NO\n"), 0;
  for (int i = 1; i <= 26; i++)
    if (tmp[i] == true && cnt[i] == 0) return printf("NO\n"), 0;
  for (int i = 1; i <= Index; i++) printf("%c", ans[i] + 'a' - 1);
  return 0;
}
int main(int argc, char** argv) {
  init();
  return 0;
}
