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
比赛的时候根本就没看懂这道题，写的什么.......。后来比完了才明白，原来是他给了你一个新定义（就是这个定意和样例没联系起来），然后让你按着这个定义排序，然后输出。我感觉这道题可难了，因为他的新定义很麻烦啊，如果上面的那个是下面的子串就没事，如果下面是上面的子串就得输出"Impossible"，然后还得使最右边的字母之间有一条边，然后拓扑排序。例如abc和cde，那么c和e之间就得弄一条边。
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[110],map[50][50],b[110];
char s[110][110],c[50];
void cn()
{
int p,flag,cns=0;
for(int i=0; i<26; i++)
{
int flag=0;
for(int j=0; j<26; j++)
{
if(!b[j])
{
p=j;
flag=1;
break;
}
}
if(!flag)
{
printf("Impossible\n");
return;
}
b[p]--;
c[cns++]=p+'a';
for(int j=0; j<26; j++)
if(map[p][j])
b[j]--;
}
c[cns]='\0';
puts(c);
}
int main()
{
int n;
scanf("%d",&n);
for(int i=0; i<n; i++)
{
scanf("%s",s[i]);
a[i]=strlen(s[i]);
}
int flag=0;
for(int i=0; i<n; i++)
{
for(int j=i+1; j<n; j++)
{
int l1=a[i];
int l2=a[j];
int k,a=0;
for(k=0; k<l1&&k<l2; k++)
{
if(s[i][k]!=s[j][k])
{
a=1;
if(!map[s[i][k]-'a'][s[j][k]-'a'])
{
map[s[i][k]-'a'][s[j][k]-'a']=1;
b[s[j][k]-'a']++;
}
break;
}
}
if(!a)
{
if(k!=l1)
{
flag=1;
break;
}
}
}
if(flag)
break;
}
if(flag)
printf("Impossible\n");
else
cn();
}