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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
const int MAXN = 100000;
const int MAXLEN = 100000;
int n;
char s[MAXLEN + 1];
int off[MAXN + 1];
int len[MAXN];
int nxt[26];
bool any[26];
bool fst[26];
bool done[26];
char ret[27];
int nret;
bool solve() {
  for (int i = (0); i < (26); ++i)
    any[i] = false, nxt[i] = -1, fst[i] = true, done[i] = false;
  for (int i = (0); i < (n); ++i)
    for (int j = (off[i]); j < (off[i + 1]); ++j) {
      int x = s[j] - 'a';
      any[x] = true;
      if (j + 1 < off[i + 1]) {
        int y = s[j + 1] - 'a';
        if (nxt[x] != -1 && nxt[x] != y) return false;
        nxt[x] = y;
        fst[y] = false;
      }
    }
  nret = 0;
  for (int i = (0); i < (26); ++i)
    if (any[i] && fst[i]) {
      int x = i;
      while (x != -1 && !done[x]) {
        ret[nret++] = 'a' + x;
        done[x] = true;
        x = nxt[x];
      }
      if (x != -1) return false;
    }
  for (int i = (0); i < (26); ++i)
    if (any[i] && !done[i]) return false;
  return true;
}
void run() {
  scanf("%d", &n);
  off[0] = 0;
  for (int i = (0); i < (n); ++i) {
    scanf("%s", s + off[i]);
    len[i] = strlen(s + off[i]);
    off[i + 1] = off[i] + len[i];
  }
  if (!solve()) {
    printf("NO\n");
    return;
  }
  printf("%s\n", ret);
}
int main() {
  run();
  return 0;
}
