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, n, i, j, k, h, x, gg, a[1111], L[1111], R[1111], in[1111], out[1111],
    vis[28], vs[28], vss[28];
string s[1111], _S;
vector<string> S;
vector<int> V[1111];
void ins(int x, int y) { V[x].push_back(y); }
void dfs(int x) {
  vis[x] = 1;
  _S.push_back('a' + x);
  vs[x] = 0;
  for (int i = 0, y; i < V[x].size(); i++)
    if (!vis[y = V[x][i]])
      dfs(y);
    else
      gg = 1;
}
int main() {
  scanf("%d", &n);
  if (n > 1000) return puts("NO"), 0;
  for (i = 1; i <= n; i++) {
    cin >> s[i];
    if (s[i].length() == 1) vs[s[i][0] - 'a'] = 1;
    vss[s[i][0] - 'a'] = 1;
    for (j = 1; j < s[i].length(); j++) {
      ins(s[i][j - 1] - 'a', s[i][j] - 'a');
      vss[s[i][j] - 'a'] = 1;
    }
  }
  for (i = 0; i < 26; i++)
    sort(V[i].begin(), V[i].end()),
        V[i].resize(unique(V[i].begin(), V[i].end()) - V[i].begin());
  for (i = 0; i < 26; i++)
    for (in[i] = V[i].size(), j = 0; j < V[i].size(); j++) out[V[i][j]]++;
  for (i = 0; i < 26; i++)
    if (V[i].size() > 1 || out[i] > 1) return puts("NO"), 0;
  for (i = 0; i < 26; i++)
    if (vss[i])
      if (in[i] && !out[i]) _S = "", dfs(i), S.push_back(_S);
  if (gg) return puts("NO"), 0;
  for (i = 0; i < 26; i++)
    if (!vis[i] && out[i]) return puts("NO"), 0;
  for (i = 0; i < 26; i++)
    if (!vis[i] && vs[i]) {
      _S = 'a' + i;
      S.push_back(_S);
    }
  sort(S.begin(), S.end());
  for (int i = 0; i < S.size(); i++) cout << S[i];
}
