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 MAXN = 210;
int n;
string mas[MAXN];
set<int> sts[MAXN];
bool Unite(int i, int j) {
  int c = mas[i][j];
  for (int k = 0; k < (int)mas[c].size(); ++k) {
    if (j + k < (int)mas[i].size()) {
      if (mas[i][j + k] != mas[c][k]) {
        return false;
      }
    } else {
      mas[i] += mas[c][k];
      if (sts[i].count(mas[c][k])) {
        return false;
      } else {
        sts[i].insert(mas[c][k]);
      }
    }
  }
  mas[c].clear();
  sts[c].clear();
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  bool fl = true;
  for (int i = 0; i < n; ++i) {
    string cur;
    cin >> cur;
    int c = cur[0];
    if (cur.size() >= mas[c].size()) {
      if (cur.substr(0, mas[c].size()) == mas[c]) {
        mas[c] = cur;
      } else {
        fl = false;
      }
    } else if (mas[c].substr(0, cur.size()) != cur) {
      fl = false;
    }
  }
  if (!fl) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < MAXN; ++i) {
    set<int> st;
    for (auto e : mas[i]) {
      if (st.count(e)) {
        cout << "NO" << endl;
        return 0;
      } else {
        st.insert(e);
      }
    }
    sts[i] = st;
  }
  for (int i = 0; i < MAXN; ++i) {
    int j = 1;
    while (j < (int)mas[i].size()) {
      if (!Unite(i, j)) {
        cout << "NO" << endl;
        return 0;
      }
      j++;
    }
  }
  string ans;
  for (int i = 0; i < MAXN; ++i) {
    ans += mas[i];
  }
  set<char> st;
  for (auto e : ans) {
    if (st.count(e)) {
      cout << "NO" << endl;
      return 0;
    } else {
      st.insert(e);
    }
  }
  cout << ans << endl;
  return 0;
}
