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;
int nxt[500];
int deg[500];
int has[500];
string ans;
int vis[500];
bool explore(int v) {
  if (vis[v]) return false;
  vis[v] = 1;
  ans += (char)v;
  if (nxt[v] == 0) return true;
  return explore(nxt[v]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  memset(vis, 0, sizeof vis);
  memset(has, 0, sizeof has);
  bool ok = true;
  for (int i = 0; i < n; i++) {
    string s;
    cin >> s;
    for (int i = 0; i < (int)s.size() - 1; i++) {
      has[s[i]] = 1;
      if (nxt[s[i]] != s[i + 1] && nxt[s[i]] != 0) {
        ok = false;
        break;
      } else {
        nxt[s[i]] = s[i + 1];
        deg[s[i + 1]]++;
      }
    }
    has[s[(int)s.size() - 1]] = 1;
    if (!ok) {
      break;
    }
  }
  if (!ok) {
    cout << "NO" << endl;
  } else {
    for (int i = 0; i < 500; i++) {
      if (has[i] && deg[i] == 0) {
        if (!explore(i)) {
          ok = false;
          break;
        }
      }
    }
    for (int i = 0; i < 500; i++) {
      if (has[i] && vis[i] == 0) {
        ok = false;
        break;
      }
    }
    if (!ok) {
      cout << "NO" << endl;
    } else {
      cout << ans << endl;
    }
  }
}
