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>
#pragma comment(linker, "/STACK:102400000,102400000")
#pragma GCC optimize("O3")
using namespace std;
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
const int maxn = 1e5 + 10;
const long long mod = 1e9 + 7;
long long fsp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans % mod;
}
int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return f * x;
}
bitset<30> vis, tp;
int nt[40], pre[40];
string s;
int main() {
  vis.reset();
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= 25; i++) nt[i] = pre[i] = -1;
  for (int i = 1; i <= n; i++) {
    cin >> s;
    tp.reset();
    for (auto c : s)
      if (tp[c - 'a'])
        return 0 * puts("NO");
      else
        tp[c - 'a'] = 1;
    vis |= tp;
    for (int j = 1; j < s.length(); j++) {
      int pv = s[j - 1] - 'a', k = s[j] - 'a';
      if (nt[pv] == k && pre[k] == pv)
        continue;
      else if (nt[pv] == -1 && pre[k] == -1)
        nt[pv] = k, pre[k] = pv;
      else
        return 0 * puts("NO");
    }
  }
  string ans = "";
  vector<int> v;
  v.clear();
  for (int i = 0; i <= 25; i++)
    if (pre[i] == -1 && vis[i]) v.push_back(i);
  for (auto it : v)
    for (int i = it; ~i; i = nt[i]) ans += 'a' + i, vis[i] = 0;
  for (int i = 0; i <= 25; i++)
    if (vis[i]) return 0 * puts("NO");
  cout << ans << endl;
  return 0;
}
