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;
bool m = true;
int n;
vector<char> a[100005];
char st[100005];
int r[26];
int pl[26];
int l[100005];
vector<pair<int, int> > s[26];
int p[26];
vector<int> d;
bool vs[26];
bool rt[26];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%s", st);
    l[i] = strlen(st);
    for (int j = 0; j < l[i]; j++) {
      a[i].push_back(st[j]);
      p[st[j] - 'a']++;
      if (p[st[j] - 'a'] >= 2)
        m = false;
      else
        s[st[j] - 'a'].push_back(make_pair(i, j));
    }
    for (int j = 0; j < 26; j++) p[j] = 0;
  }
  if (!m) {
    printf("NO");
    return 0;
  }
  int u, v;
  int nl;
  int cl;
  for (int i = 0; i < 26; i++) {
    r[i] = -1;
    pl[i] = -1;
  }
  for (int i = 0; i < 26; i++) {
    cl = -1;
    for (int j = 0; j < s[i].size(); j++) {
      rt[i] = true;
      u = s[i][j].first;
      v = s[i][j].second;
      if (l[u] != v + 1) {
        nl = a[u][v + 1] - 'a';
        if (!((r[nl] >= 0 && r[nl] != i) || (cl >= 0 && cl != nl))) {
          r[nl] = i;
          pl[i] = nl;
          cl = nl;
        } else {
          printf("NO");
          return 0;
        }
      }
    }
  }
  int k;
  for (int i = 0; i < 26; i++) {
    u = i;
    k = 0;
    if (rt[i]) {
      while (u != -1) {
        k++;
        if (k > 27) {
          printf("NO");
          return 0;
        }
        if (r[u] == -1 && !vs[u]) d.push_back(u);
        vs[u] = true;
        u = r[u];
      }
    }
  }
  sort(d.begin(), d.end());
  char sl;
  for (int i = 0; i < d.size(); i++) {
    u = d[i];
    while (u != -1) {
      sl = u + 'a';
      cout << sl;
      u = pl[u];
    }
  }
  cout << endl;
  return 0;
}
