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;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
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();
            char[] next = new char[256];
            char[] prev = new char[256];
            boolean[] need = new boolean[256];
            Arrays.fill(next, '@');
            Arrays.fill(prev, '@');
            while (N > 0) {
                //out.println(N);
                String str = in.next();
                if (str.length() > 26) {
                    out.println("NO");
                    return;
                }
                if (str.length() == 1) {
                    need[str.charAt(0)] = true;
                    N--;
                    continue;
                }
                for (int i = 0; i < str.length(); i++) {
                    for (int j = i + 1; j < str.length(); j++) {
                        if (str.charAt(i) == str.charAt(j)) {
                            out.println("NO");
                            return;
                        }
                    }
                }
                for (int j = 0; j < str.length() - 1; j++) {
                    char from = str.charAt(j);
                    char to = str.charAt(j + 1);
                    if (next[from] != '@' && next[from] != to) {
                        out.println("NO");
                        return;
                    }
                    if (prev[to] != '@' && prev[to] != from) {
                        out.println("NO");
                        return;
                    }
                    for (char i = to; i != '@'; i = next[i]) {
                        if (i == from) {
                            out.println("NO");
                            return;
                        }
                    }
                    for (int i = from; i != '@'; i = prev[i]) {
                        if (i == to) {
                            out.println("NO");
                            return;
                        }
                    }
                    next[from] = to;
                    prev[to] = from;
                }
                N--;
            }
            for (char i = 'a'; i <= 'z'; i++) {
                if ((next[i] != '@' && prev[i] == '@') || (need[i] && prev[i] == '@' && next[i] == '@')) {
                    for (char j = i; j != '@'; j = next[j]) out.print(j);
                }
            }

        }

    }
}

