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;
char str[100010];
int n = 0;
int nexT[27] = {0};
int last[27] = {0};
int number = 0;
int start[27] = {0};
int len = 0;
char s[100010];
bool isUsed[27] = {false};
int main(int argc, const char* argv[]) {
  cin >> n;
  bool abort = false;
  for (int i = 0; i < n; ++i) {
    scanf("%s", str);
    if (abort) {
      continue;
    }
    for (int i = 0; i <= strlen(str); ++i) {
      if (i == 0) {
        if (last[str[i] - 'a' + 1] == 0) {
          last[str[i] - 'a' + 1] = -1;
        }
        continue;
      }
      isUsed[str[i - 1] - 'a' + 1] = true;
      if (i == strlen(str)) {
        continue;
      }
      if (nexT[str[i - 1] - 'a' + 1] == 0 && last[str[i] - 'a' + 1] != 1) {
        last[str[i] - 'a' + 1] = 1;
        nexT[str[i - 1] - 'a' + 1] = str[i] - 'a' + 1;
      } else if (nexT[str[i - 1] - 'a' + 1] == str[i] - 'a' + 1) {
        continue;
      } else {
        abort = true;
        break;
      }
    }
  }
  for (int i = 1; i <= 26; ++i) {
    if (last[i] == -1) {
      start[number++] = i;
    }
  }
  if (abort) {
    cout << "NO" << endl;
  } else {
    for (int i = 0; i < number; ++i) {
      int pos = start[i];
      s[len++] = pos + 'a' - 1;
      isUsed[pos] = false;
      while (nexT[pos] != 0) {
        pos = nexT[pos];
        isUsed[pos] = false;
        s[len++] = pos + 'a' - 1;
      }
    }
    s[len++] = '\0';
    for (int i = 1; i <= 26; ++i) {
      if (isUsed[i]) {
        abort = true;
      }
    }
    if (abort) {
      cout << "NO" << endl;
    } else {
      cout << s << endl;
    }
  }
}
