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 std::sort;
using std::vector;
int cmp(const void *x, const void *y) {
  return (*((double *)(x))) > (*((double *)(y))) ? 1 : -1;
}
bool v[300], g[300];
int next[300], pre[300];
char str[100005];
bool dfs(int x) {
  if (g[x]) return false;
  g[x] = true;
  if (next[x] == 0) return true;
  return dfs(next[x]);
}
int main() {
  int n, len;
  while (scanf("%d", &n) != EOF) {
    memset(next, 0, sizeof(next));
    memset(pre, 0, sizeof(pre));
    memset(v, false, sizeof(v));
    bool flag = true;
    for (int i = 0; i < n; i++) {
      scanf("%s", str);
      len = strlen(str);
      for (int j = 0; j < len - 1; j++) {
        v[str[j]] = true;
        if (next[str[j]] == 0)
          next[str[j]] = str[j + 1];
        else if (next[str[j]] != str[j + 1])
          flag = false;
        if (pre[str[j + 1]] == 0)
          pre[str[j + 1]] = str[j];
        else if (pre[str[j + 1]] != str[j])
          flag = false;
      }
      v[str[len - 1]] = true;
    }
    for (int i = 'a'; i <= 'z'; i++) {
      memset(g, false, sizeof(g));
      if (!dfs(i)) flag = false;
    }
    if (!flag)
      printf("NO\n");
    else {
      len = 0;
      for (int i = 'a'; i <= 'z'; i++) {
        if (v[i]) {
          bool root = true;
          for (int j = 'a'; j <= 'z'; j++)
            if (next[j] == i) root = false;
          if (!root) continue;
          int ptr = i;
          while (ptr != 0) {
            str[len++] = ptr;
            v[ptr] = false;
            ptr = next[ptr];
          }
        }
      }
      str[len] = '\0';
      printf("%s\n", str);
    }
  }
  return 0;
}
