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;
mt19937_64 rng((unsigned long long)new char);
const double eps = 1e-7;
const int maxn = 1e6 + 100;
const long long INF = 9e8;
const int mod = 1e9 + 7;
bool g[26][26];
string s[maxn];
int n;
bool vis[26];
bool done[26];
bool dfs(int u) {
  done[u] = 1;
  for (int t = (0); t < (26); ++t) {
    if (g[u][t]) {
      if (done[t]) return 0;
      if (!dfs(t)) return 0;
    }
  }
  return 1;
}
string ans;
void dfs2(int u) {
  done[u] = 1;
  ans += u + 'a';
  for (int t = (0); t < (26); ++t) {
    if (g[u][t]) {
      dfs2(t);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int t = (0); t < (n); ++t) {
    cin >> s[t];
    for (int i = (0); i < (s[t].length() - 1); ++i) {
      g[s[t][i] - 'a'][s[t][i + 1] - 'a'] = 1;
    }
    for (auto& c : s[t]) vis[c - 'a'] = 1;
  }
  bool fl = 1;
  for (int t = (0); t < (26); ++t) {
    int cnt = 0;
    for (int j = (0); j < (26); ++j) cnt += g[t][j];
    if (cnt >= 2) {
      fl = 0;
      break;
    }
  }
  for (int t = (0); t < (26); ++t) {
    int cnt = 0;
    for (int j = (0); j < (26); ++j) cnt += g[j][t];
    if (cnt >= 2) {
      fl = 0;
      break;
    }
  }
  for (int t = (0); t < (26); ++t)
    if (vis[t]) {
      memset(done, 0, sizeof(done));
      if (!dfs(t)) {
        fl = 0;
        break;
      }
    }
  if (fl) {
    memset(done, 0, sizeof(done));
    int in[26];
    memset(in, 0, sizeof(in));
    for (int t = (0); t < (26); ++t) {
      for (int i = (0); i < (26); ++i) in[t] += g[i][t];
    }
    for (int t = (0); t < (26); ++t) {
      if (vis[t] && !done[t] && !in[t]) {
        dfs2(t);
      }
    }
    cout << ans;
  } else
    cout << "NO";
}
