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;
inline int read(register int ans = 0, register int sgn = ' ',
                register int ch = getchar()) {
  for (; ch < '0' || ch > '9'; sgn = ch, ch = getchar())
    ;
  for (; ch >= '0' && ch <= '9'; (ans *= 10) += ch - '0', ch = getchar())
    ;
  return sgn - '-' ? ans : -ans;
}
char _Ob[1000000 + 5], *_O = _Ob, _Os[25], *_Ot;
template <typename T>
inline void Out(T x) {
  if (!x) {
    (*_O++ = '0');
    return;
  }
  if (x < 0) (*_O++ = '-'), x = -x;
  for (_Ot = _Os; x; *_Ot++ = x % 10 + '0', x /= 10)
    ;
  for (; _Ot != _Os; (*_O++ = *--_Ot))
    ;
  if (_O - _Ob >= 1000000 - 50) fwrite(_Ob, 1, _O - _Ob, stdout), _O = _Ob;
}
char s[100000 + 5], ans[27];
bool o[27];
int in[27], fa[27], n, m, c;
void DFS(int j) {
  ans[++c] = j + 'a' - 1, o[j] = false;
  if (fa[j] && o[fa[j]]) --in[fa[j]], DFS(fa[j]);
}
int main() {
  n = read();
  for (register int i = 1, j; i <= n; i++) {
    for (scanf("%s", s + 1), m = strlen(s + 1), j = 1; j <= m;
         o[s[j] -= 'a' - 1] = true, j++)
      ;
    for (j = 1; j < m; j++)
      if (fa[s[j]] && fa[s[j]] != s[j + 1]) {
        puts("NO");
        return 0;
      } else
        in[s[j + 1]] += !fa[s[j]], fa[s[j]] = s[j + 1];
  }
  for (register int j = 1; j <= 26; j++)
    if (in[j] > 1) {
      puts("NO");
      return 0;
    }
  for (register int j = 1; j <= 26; o[j] && !in[j] ? DFS(j), 0 : 0, j++)
    ;
  for (register int j = 1; j <= 26; j++)
    if (o[j] && in[j]) {
      puts("NO");
      return 0;
    }
  printf("%s\n", ans + 1);
  return 0;
}
