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 dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
string tostring(long long int res) {
  string curstr = "";
  while (res != 0) {
    long long int temp = (res % 10);
    curstr += ((char)temp + '0');
    res /= 10;
  }
  reverse(curstr.begin(), curstr.end());
  return curstr;
}
long long int toint(string ss) {
  long long int inss = 0;
  for (int i = 0; i < ss.size(); i++) {
    inss = (inss * 10) + ((int)(ss[i] - '0'));
  }
  return inss;
}
vector<int> graph[27];
int cycle, fl = 1, vis[29], ache[29], deg[29], vis2[27][27];
string curstr;
void dfs(int cur, int kontateachi) {
  if (cycle) return;
  vis2[kontateachi][cur] = 1;
  for (int i = 0; i < graph[cur].size(); i++) {
    int vv = graph[cur][0];
    if (!vis2[kontateachi][vv])
      dfs(vv, kontateachi);
    else {
      cycle = 1;
      return;
    }
  }
}
void dfs2(int cur) {
  vis[cur] = 1;
  if (!fl) return;
  for (int i = 0; i < graph[cur].size(); i++) {
    int vv = graph[cur][i];
    if (vis[vv]) {
      fl = 0;
      return;
    }
    if (!vis[vv]) curstr += (char)(vv + 'a'), dfs2(vv);
  }
}
int main() {
  int n;
  string str, res;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    cin >> str;
    int len = str.length();
    for (int i = 0; i < len; i++) ache[str[i] - 'a'] = 1;
    for (int i = 0; i < len - 1; i++) {
      int u = str[i] - 'a', v = str[i + 1] - 'a';
      if ((count(graph[u].begin(), graph[u].end(), v)) == 0)
        graph[u].push_back(v);
      deg[v]++;
    }
  }
  for (int i = 0; i < 26; i++) {
    if (!ache[i]) continue;
    if (graph[i].size() > 1) {
      fl = 0;
      break;
    }
    dfs(i, i);
    if (cycle) {
      fl = 0;
      break;
    }
  }
  if (fl) {
    set<string> ss;
    for (int i = 0; i < 26; i++) {
      if (!vis[i] && ache[i] && !deg[i]) {
        curstr = (char)(i + 'a');
        dfs2(i);
        ss.insert(curstr);
      }
    }
    res = "";
    if (fl) {
      set<string>::iterator it;
      for (it = ss.begin(); it != ss.end(); it++) res += *it;
      cout << res << endl;
    } else
      cout << "NO" << endl;
  } else
    cout << "NO" << endl;
  return 0;
}
