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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e5 + 5;
char s[N];
int pre[26], nxt[26], n;
bool vis[26], vv[26];
int main() {
  memset(pre, -1, sizeof(pre)), memset(nxt, -1, sizeof(nxt));
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%s", s);
    int l = strlen(s);
    for (int j = 0; j < l; ++j) vis[s[j] - 'a'] = true;
    for (int j = 0; j + 1 < l; ++j) {
      int x = s[j] - 'a', y = s[j + 1] - 'a';
      if (nxt[x] != -1 && nxt[x] != y) return 0, puts("NO");
      if (pre[y] != -1 && pre[y] != x) return 0, puts("NO");
      nxt[x] = y;
      pre[y] = x;
    }
  }
  for (int i = 0; i < 26; ++i) {
    if (vis[i] && pre[i] == -1) {
      int x = i;
      while (true) {
        vv[x] = true;
        x = nxt[x];
        if (x == -1) break;
      }
    }
  }
  for (int i = 0; i < 26; ++i)
    if (vis[i] && !vv[i]) return 0, puts("NO");
  for (int i = 0; i < 26; ++i) {
    if (vis[i] && pre[i] == -1) {
      int x = i;
      while (true) {
        putchar('a' + x);
        x = nxt[x];
        if (x == -1) break;
      }
    }
  }
  return 0;
}
