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.
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int[] prev = new int[26];
            int[] next = new int[26];
            int[] cnt = new int[26];
            Arrays.fill(prev, -1);
            Arrays.fill(next, -1);
            Arrays.fill(cnt, 0);
            for (int i = 0; i < n; ++i) {
                String s = in.next();
                for (int j = 1; j < s.length(); ++j) {
                    int x = s.charAt(j - 1) - 'a';
                    int y = s.charAt(j) - 'a';
                    if (next[x] != -1 && next[x] != y || prev[y] != -1 && prev[y] != x) {
                        out.println("NO");
                        return;
                    }
                    next[x] = y;
                    prev[y] = x;
//                System.out.printf("%d %d \n", x, y);
                }
                for (int j = 0; j < s.length(); j++) {
                    cnt[s.charAt(j) - 'a'] = 1;
                }
            }
            String ans = "";
            for (int k = 0; k < 26; ++k) {
                if (cnt[k] != 0 && prev[k] == -1) {
//                out.println((char)(k + 'a'));
                    for (int j = k; j != -1; j = next[j]) {
                        --cnt[j];
                        ans += (char) (j + 'a');
                    }
                }
            }
            for (int k = 0; k < 26; ++k) {
                if (cnt[k] != 0) {
                    out.println("NO");
                    return;
                }
            }
            out.println(ans);
        }

    }
}

