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 = 1e5 + 10;
string s[maxn];
vector<int> G[55];
int r[55] = {0};
int c[55] = {0};
bool vis[55] = {0};
char res[105][105];
int cnt = 0;
int tot = 0;
int temp[55] = {0};
string a[30];
void dfs(int u, int cur) {
  res[tot][cur] = u + 'a';
  int n = G[u].size();
  if (!n) {
    return;
  } else {
    dfs(G[u][0], cur + 1);
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
  }
  for (int i = 1; i <= n; ++i) {
    int m = s[i].size();
    for (int j = 0; j < m; ++j) {
      if (!vis[s[i][j] - 'a']) {
        vis[s[i][j] - 'a'] = 1;
        cnt++;
      }
    }
    for (int j = 0; j < m - 1; ++j) {
      int k = G[s[i][j] - 'a'].size();
      if (!k) {
        G[s[i][j] - 'a'].push_back(s[i][j + 1] - 'a');
        c[s[i][j] - 'a']++;
        r[s[i][j + 1] - 'a']++;
        temp[s[i][j + 1] - 'a']++;
      } else {
        if (G[s[i][j] - 'a'][0] != s[i][j + 1] - 'a') {
          cout << "NO" << endl;
          return 0;
        }
      }
    }
  }
  for (int i = 0; i < 30; ++i) {
    if (r[i] > 1 || c[i] > 1) {
      cout << "NO" << endl;
      return 0;
    }
  }
  queue<int> q;
  for (int i = 0; i < 30; ++i) {
    if (vis[i] && r[i] == 0) {
      q.push(i);
      cnt--;
    }
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    int m = G[u].size();
    for (int i = 0; i < m; ++i) {
      r[G[u][i]]--;
      if (r[G[u][i]] == 0) {
        q.push(G[u][i]);
        cnt--;
      }
    }
  }
  if (cnt) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < 30; ++i) {
    if (vis[i] && temp[i] == 0) {
      dfs(i, 0);
      tot++;
    }
  }
  for (int i = 0; i < tot; ++i) {
    a[i] = res[i];
  }
  sort(a, a + tot);
  for (int i = 0; i < tot; ++i) {
    cout << a[i];
  }
  cout << endl;
  return 0;
}
