Problem: Fox Ciel is going to publish a paper on FOCS (Foxes Operated Computer Systems, pronounce: "Fox"). She heard a rumor: the authors list on the paper is always sorted in the lexicographical order. 

After checking some examples, she found out that sometimes it wasn't true. On some papers authors' names weren't sorted in lexicographical order in normal sense. But it was always true that after some modification of the order of letters in alphabet, the order of authors becomes lexicographical!

She wants to know, if there exists an order of letters in Latin alphabet such that the names on the paper she is submitting are following in the lexicographical order. If so, you should find out any such order.

Lexicographical order is defined in following way. When we compare s and t, first we find the leftmost position with differing characters: si ≠ ti. If there is no such position (i. e. s is a prefix of t or vice versa) the shortest string is less. Otherwise, we compare characters si and ti according to their order in alphabet.

Input

The first line contains an integer n (1 ≤ n ≤ 100): number of names.

Each of the following n lines contain one string namei (1 ≤ |namei| ≤ 100), the i-th name. Each name contains only lowercase Latin letters. All names are different.

Output

If there exists such order of letters that the given names are sorted lexicographically, output any such order as a permutation of characters 'a'–'z' (i. e. first output the first letter of the modified alphabet, then the second, and so on).

Otherwise output a single word "Impossible" (without quotes).

Examples

Input

3
rivest
shamir
adleman


Output

bcdefghijklmnopqrsatuvwxyz


Input

10
tourist
petr
wjmzbmr
yeputons
vepifanov
scottwu
oooooooooooooooo
subscriber
rowdark
tankengineer


Output

Impossible


Input

10
petr
egor
endagorion
feferivan
ilovetanyaromanova
kostka
dmitriyh
maratsnowbear
bredorjaguarturnik
cgyforever


Output

aghjlnopefikdmbcqrstuvwxyz


Input

7
car
care
careful
carefully
becarefuldontforgetsomething
otherwiseyouwillbehacked
goodluck


Output

acbdefhijklmnogpqrstuvwxyz

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;
import java.util.Scanner;

public class Test
{   
    // 输入的名字
    private static String[] names = new String[200];
    // 建立的图
    private static boolean[][] graph = new boolean[30][30];
    // 计算每个点的入度
    private static int[] inDegree = new int[30];
    // 提前判断
    private static boolean result = true;
    
    // srm 279
    public static void main(String[] args)
    {       
        Scanner scan = new Scanner(System.in);
        int i , n = scan.nextInt();
        for (i = 0;i < n;i ++)
            names[i] = scan.next();
        
        buildGraph(n);
        if (!result)
            System.out.println("Impossible");
        else
            solve();
    }
    
    // 根据名字建立图
    private static void buildGraph(int n)
    {
        int i , j;
        for (i = 0;i < n;i ++)
            for (j = i + 1;j < n;j ++)
                update(names[i] , names[j]);
    }
    
    // 找到第一个不一样的，进行图的更新
    private static void update(String s1 , String s2)
    {
        int i = 0 , l1 = s1.length() , l2 = s2.length() , l = Math.min(l1 , l2);
        while (i < l && (s1.charAt(i) == s2.charAt(i)))
            i ++;
        // 如果找到了不一样的
        if (i < l)
        {
            char ch1 = s1.charAt(i);
            char ch2 = s2.charAt(i);
            // 只有在第一次进行++
            if (!graph[toInt(ch1)][toInt(ch2)])
                inDegree[toInt(ch2)] ++;
            graph[toInt(ch1)][toInt(ch2)] = true;
        }
        // 如果前面都一样
        else
        {
            if (l1 > l2)
                result = false;
        }
    }
    
    // 进行拓扑排序的求解
    private static void solve()
    { 
        int cot = 0;
        List<Character> outputList = new ArrayList<Character>();
        // 存储入度为0的点
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i = 1;i <= 26;i ++)
        {
            // 如果入度为0
            if (inDegree[i] == 0)
                queue.add(i);
        }
        
        // 直到全部处理完
        while (!queue.isEmpty())
        {
            int node = queue.poll();
            outputList.add(toChar(node));
            cot ++;
            for (int i = 1;i <= 26;i ++)
            {
                if (graph[node][i])
                {
                    inDegree[i] --;
                    if (inDegree[i] == 0)
                        queue.add(i);
                }
            }
        }
        if (cot == 26)
        {
            for (Character ch : outputList)
                System.out.print(ch);
            System.out.println();
        }
        else
            System.out.println("Impossible");
    }
    
    private static int toInt(char ch)
    {
        return ((ch - 'a') + 1);
    }
    
    private static char toChar(int index)
    {
        return (char) ((index - 1) + 'a');
    }
    
}