Codeforces Round #290 (Div. 2) C. Fox and Names topological ordering

C. Fox and Namestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

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 are always sorted in the lexicographical order.

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

She wants to know, if there exists a order of letters in Latin alphabet such that the names on the paper she is Submittin G is following in the lexicographical order. If So, you should find off any such order.

lexicographicalOrder is defined in following. When we comparesandT, first we find the leftmost position with differing characters: si? ≠? T I . If There is no such position (I. E.sis a prefix ofTor 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 followingNLines contain one string namei (1?≤?| name I|? ≤?100), theI-th name. Each name contains only lowercase Latin letters. All names is different.

Output

If there exists such order of letters that the given names is sorted lexicographically, output any such order as a Permut ation 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).

Sample Test (s) input

3rivestshamiradleman

Output

Bcdefghijklmnopqrsatuvwxyz

Input

10touristpetrwjmzbmryeputonsvepifanovscottwuoooooooooooooooosubscriberrowdarktankengineer

Output

Impossible

Input

10petregorendagorionfeferivanilovetanyaromanovakostkadmitriyhmaratsnowbearbredorjaguarturnikcgyforever

Output

Aghjlnopefikdmbcqrstuvwxyz

Input

7carcarecarefulcarefullybecarefuldontforgetsomethingotherwiseyouwillbehackedgoodluck

Output

Acbdefhijklmnogpqrstuvwxyz

The topic gives a string set, and assumes that it is a dictionary order, requires 26-Letter dictionary (custom dictionary order), every two strings, you can derive the precedence of a pair of characters, and then the order of all the characters, then transformed into a topological sorting problem, using the priority queue, This can be as small as possible in front. Second, if there is ab a, such a string, can be directly considered impossible to exist!

#define N 105#define MOD 1000000000000000007struct node{int x;    Node (int xx) {x = xx;    } BOOL operator < (const Node A) const{return x>a.x; }};int N,in[30],ans[30],ansnum;char str[n][n];bool land[30][30];p riority_queue<node> myqueue;bool getland (int    X,int y) {int len = min (strlen (str[x]), strlen (Str[y]));            FI (len) {if (Str[x][i]! = Str[y][i]) {Land[str[x][i]-' A '][str[y][i]-' a '] = true;        return true;    }} if (Strlen (Str[x]) >strlen (Str[y])) return false; return true;}        int main () {while (S (n)!=eof) {FI (n) {SS (str[i]);        } memset (Land,false,sizeof (land));        BOOL flag = TRUE; for (int i=0;i<n && flag;i++) {for (int j = i+1;j<n && flag;j++) {flag = Get            Land (I,J);            }} if (!flag) {printf ("impossible\n");        Continue        } memset (In,0,sizeof (in)); FI (26) {FJ (+) {if (Land[i][j]) {in[j]++;        }}} while (!myqueue.empty ()) Myqueue.pop ();            for (int i=0;i<26;i++) {if (in[i] = = 0) {Myqueue.push (node (i));        }} ansnum = 0;            while (!myqueue.empty ()) {Node top = myqueue.top ();            Myqueue.pop ();            ans[ansnum++] = top.x;                    FI (+) {if (Land[top.x][i]) {in[i]--;                if (in[i] = = 0) Myqueue.push (node (i));            }}} if (Ansnum = =) {FI (ansnum) printf ("%c", ans[i]+ ' a ');        printf ("\ n");    } else printf ("impossible\n"); } return 0;}

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Codeforces Round #290 (Div. 2) C. Fox and Names topological ordering