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;
const int N = 1e5 + 5;
int n, len, num, tot[30], x[30], y[30];
bool f[30], ok[30], chu[30];
char s[N], ans[30][30], tmp[30];
void end() {
  printf("NO");
  exit(0);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%s", s + 1);
    len = strlen(s + 1);
    memset(f, 0, sizeof(f));
    memset(chu, 0, sizeof(chu));
    for (int j = 1; j <= len; j++) {
      if (f[s[j] - 'a']) end();
      f[s[j] - 'a'] = true;
    }
    int noc = 0, cntt = 0, last = 1, qaq = 0;
    for (int j = 1; j <= len; j++) {
      int t = s[j] - 'a';
      if (!y[t]) {
        noc = j;
        continue;
      }
      if (j != 1 && y[t] != 1) end();
      int id = x[t], now = j;
      if (chu[id]) end();
      if (!qaq)
        qaq = id;
      else
        ok[id] = true;
      chu[id] = true;
      for (int k = y[t]; k <= tot[id] && j + k - y[t] <= len; k++) {
        if (ans[id][k] != s[j + k - y[t]]) end();
        now = j + k - y[t] + 1;
      }
      for (int k = last; k <= noc; k++) tmp[++cntt] = s[k];
      for (int k = 1; k <= tot[id]; k++) tmp[++cntt] = ans[id][k];
      j = noc = now - 1;
      last = now;
    }
    for (int k = last; k <= len; k++) tmp[++cntt] = s[k];
    tot[qaq] = cntt;
    for (int k = 1; k <= cntt; k++) {
      ans[qaq][k] = tmp[k];
      x[tmp[k] - 'a'] = qaq;
      y[tmp[k] - 'a'] = k;
    }
    if (!qaq) {
      num++;
      tot[num] = len;
      for (int j = 1; j <= len; j++) {
        ans[num][j] = s[j];
        x[s[j] - 'a'] = num;
        y[s[j] - 'a'] = j;
      }
    }
  }
  int mn = -1;
  for (int i = 1; i <= num; i++) {
    int now = 26, id = 0;
    for (int j = 1; j <= num; j++)
      if (!ok[j] && ans[j][1] - 'a' > mn && ans[j][1] - 'a' < now)
        now = ans[j][1] - 'a', id = j;
    if (!id) break;
    mn = now;
    for (int j = 1; j <= tot[id]; j++) putchar(ans[id][j]);
  }
  return 0;
}
