CodeForces - 510C（拓扑排序）

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 e
4000
xists 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).

Example

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

改变26个字母的顺序使得给出的字符串是按字典序排列下来的，i+1字符串的j位置的字母a如果不等于i字符串j位置的字母b，那么字母a的顺序就得在b的后面，转化成图的话就是b有一条有向边连到a（a的拓扑顺序在b后），如果最终处理完所有字符串后建立的图出现了环（拓扑排序后有的点不能加入答案队列）或者字符串x比字符串y长且x的前leny位与y相同，即为impossible，否则输出拓扑排序后的结果。

#include<algorithm>
#include<string>
#include<iostream>
#include<stack>
#include<vector>
#include<queue>
using namespace std;
const int N = 200;
string now,pre;
int deg[26],ans[26];
vector<int> g[26];
queue<int> que;
int main()
{
int n,tot = 0;
cin >> n;
for(int i=0; i<n; i++)
{
cin >> now;
int len = min(now.length(),pre.length());
bool ok = false;
for(int j=0; j<len; j++)
{
if(now[j] != pre[j])
{
ok = true;
g[pre[j]-'a'].push_back(now[j]-'a');
deg[now[j]-'a']++;
break;
}
}
if(!ok&&now.length() < pre.length())
{
cout << "Impossible" << endl;
return 0;
}
pre=now;
}
for(int i=0; i<26; i++)
if(!deg[i])
que.push(i);
while(!que.empty())
{
int x = que.front();
ans[tot++] = x;
que.pop();
for(int i=0; i<g[x].size(); i++)
{
if(deg[g[x][i]] == 1)
que.push(g[x][i]);
if(deg[g[x][i]] != 0)
deg[g[x][i]]--;
}
}
if(tot != 26)
cout << "Impossible" << endl;
else
{
for(int i=0; i<26; i++)
{
cout <<char('a'+ ans[i]);
}
cout << endl;
}
return 0;
}