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 N;
bool has[30];
bool ghas[30];
int pos[30];
int ri[30], lf[30];
string str;
bool first;
bool vis[30];
vector<int> ans;
bool con(int a, int b) {
  if (ri[a] && lf[b]) {
    return (ri[a] == b);
  } else if (!ri[a] && !lf[b]) {
    ri[a] = b;
    lf[b] = a;
    return true;
  }
  return false;
}
void dfs(int u) {
  vis[u] = true;
  ans.push_back(u);
  int v = ri[u];
  if (!v)
    return;
  else if (vis[v]) {
    first = true;
    return;
  }
  dfs(v);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 1; i <= N; i++) {
    cin >> str;
    memset(has, 0, sizeof(has));
    for (char c : str) {
      int ai = c - 'a' + 1;
      if (has[ai]) {
        first = true;
        break;
      }
      has[ai] = true;
      ghas[ai] = true;
    }
    if (first) break;
    int S = str.size();
    if (S == 1) continue;
    int nx, ai;
    for (int i = 0; i <= S - 2; i++) {
      ai = str[i] - 'a' + 1;
      nx = str[i + 1] - 'a' + 1;
      if (!con(ai, nx)) {
        first = true;
        break;
      }
    }
    if (first) break;
    if (ri[nx]) {
      if (!con(nx, ri[nx])) {
        first = true;
        break;
      }
    }
  }
  if (first) {
    cout << "NO" << '\n';
    return 0;
  }
  for (int i = 1; i <= 26; i++)
    if (ghas[i] && !lf[i]) dfs(i);
  if (first) {
    cout << "NO" << '\n';
    return 0;
  }
  for (int i = 1; i <= 26; i++)
    if (ghas[i] && !vis[i]) first = true;
  if (first) {
    cout << "NO" << '\n';
    return 0;
  }
  for (int ai : ans) cout << (char)('a' + ai - 1);
  cout << '\n';
  return 0;
}
