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 N = 1e5 + 10, K = 26;
string s[N];
int g[K], p[K];
bool done[K];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> s[i];
  }
  for (int i = 0; i < K; ++i) {
    g[i] = p[i] = -1;
  }
  bool ans = true;
  for (int i = 1; i <= n and ans; ++i) {
    int m = s[i].length();
    int last = s[i][m - 1] - 'a';
    if (g[last] == -1) {
      g[last] = -2;
    }
    for (int j = 0; j < m - 1 and ans; ++j) {
      int u = s[i][j] - 'a', v = s[i][j + 1] - 'a';
      if (g[u] >= 0 and g[u] != v) {
        ans = false;
      } else if (p[v] >= 0 and p[v] != u) {
        ans = false;
      } else if (u == v) {
        ans = false;
      } else {
        g[u] = v;
        p[v] = u;
      }
    }
  }
  for (int i = 0; i < K and ans; ++i) {
    if (g[i] != -1) {
      int j = g[i];
      while (j >= 0) {
        if (j == i) {
          ans = false;
          break;
        } else {
          j = g[j];
        }
      }
    }
  }
  if (!ans) {
    cout << "NO\n";
  } else {
    for (int i = 0; i < K; ++i) {
      if (done[i] or g[i] == -1 or p[i] >= 0) {
        continue;
      } else {
        int j = i;
        while (j >= 0) {
          cout << char(j + 'a');
          done[j] = true;
          j = g[j];
        }
      }
    }
  }
}
