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;
template <class A, class B>
inline bool mina(A &first, B second) {
  return (first > second) ? (first = second, 1) : 0;
}
template <class A, class B>
inline bool maxa(A &first, B second) {
  return (first < second) ? (first = second, 1) : 0;
}
const int MAXN = 1e5 + 5;
int N;
string in[MAXN];
int nxt[26];
int pre[26];
bool have[26];
bool contains[26];
int main() {
  memset(nxt, -1, sizeof(nxt));
  memset(pre, -1, sizeof(pre));
  cin >> N;
  for (int i = 0; (i) < (N); ++(i)) {
    cin >> in[i];
    for (int j = 0; (j) < (((int)(in[i]).size())); ++(j))
      have[in[i][j] - 'a'] = true;
    for (int j = 0; (j) < (((int)(in[i]).size()) - 1); ++(j)) {
      int u = in[i][j] - 'a';
      int v = in[i][j + 1] - 'a';
      if (pre[v] != -1 && pre[v] != u) {
        cout << "NO" << endl;
        return 0;
      }
      if (nxt[u] != -1 && nxt[u] != v) {
        cout << "NO" << endl;
        return 0;
      }
      nxt[u] = v;
      pre[v] = u;
    }
  }
  vector<string> ans;
  for (int i = 0; (i) < (26); ++(i)) {
    if (nxt[i] != -1 && pre[i] == -1) {
      int now = i;
      string temp = "";
      temp += 'a' + now;
      while (nxt[now] != -1) {
        now = nxt[now];
        temp += 'a' + now;
      }
      ans.push_back(temp);
    } else if (nxt[i] == -1 && pre[i] == -1 && have[i]) {
      string temp = "";
      temp += 'a' + i;
      ans.push_back(temp);
    }
  }
  sort(ans.begin(), ans.end());
  string ans2 = "";
  for (auto s : ans) {
    ans2 += s;
  }
  for (char c : ans2) {
    contains[c - 'a'] = true;
  }
  for (int i = 0; (i) < (26); ++(i)) {
    if (have[i] && !contains[i]) {
      cout << "NO" << endl;
      return 0;
    }
  }
  cout << ans2 << endl;
  return 0;
}
