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>
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
using namespace std;
int n;
int cnt[26], c0;
bool vis[26];
vector<int> fr[26], bk[26];
string in[100005];
vector<int> ans;
int main() {
  while (~scanf("%d", &n)) {
    c0 = 0;
    memset(vis, false, sizeof(vis));
    memset(cnt, 0, sizeof(cnt));
    for (int i = 0; i <= 25; i++) fr[i].clear(), bk[i].clear();
    for (int i = 1; i <= n; i++) cin >> in[i];
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= (int)in[i].length() - 1; j++) {
        int f, b;
        f = in[i][j - 1] - 'a';
        b = in[i][j] - 'a';
        fr[b].push_back(f);
        bk[f].push_back(b);
      }
    for (int i = 1; i <= n; i++)
      for (int j = 0; j <= (int)in[i].length() - 1; j++) {
        int c = in[i][j] - 'a';
        cnt[c]++;
        if (cnt[c] == 1) c0++;
      }
    for (int i = 0; i <= 25; i++) {
      sort(fr[i].begin(), fr[i].end());
      fr[i].resize(unique(fr[i].begin(), fr[i].end()) - fr[i].begin());
      sort(bk[i].begin(), bk[i].end());
      bk[i].resize(unique(bk[i].begin(), bk[i].end()) - bk[i].begin());
    }
    bool fail = false;
    for (int i = 0; i <= 25; i++)
      if (bk[i].size() > 1) fail = true;
    if (fail) {
      puts("NO");
      continue;
    }
    ans.clear();
    while (1) {
      int cur = -1;
      for (int i = 0; i <= 25; i++)
        if (!vis[i] && cnt[i] && !fr[i].size()) {
          cur = i;
          ans.push_back(cur);
          vis[cur] = true;
          break;
        }
      if (cur == -1) break;
      while (bk[cur].size()) {
        int nxt = bk[cur][0];
        if (vis[nxt]) {
          fail = true;
          break;
        }
        ans.push_back(nxt);
        vis[nxt] = true;
        cur = nxt;
      }
    }
    if (fail || c0 != (int)ans.size()) {
      puts("NO");
      continue;
    } else {
      for (int i : ans) putchar('a' + i);
      puts("");
    }
  }
  return 0;
}
