ICPC 2012 · Problem E · Infiltration

36th ICPC · Warsaw, Poland

Statement

Problem ID: infiltration

Good morning, agent W-12. Your mission, should you choose to accept it, is as follows.

We are infiltrating the ever so insidious Association of Chaos and Mischief (ACM) in order to take down their command structure. Unfortunately, they appear to be prepared for such an eventuality, and have given their command structure an annoyingly complex design which makes our infiltration quite difficult.

The ACM command structure is divided into several cells. For each pair of cells A and B, either A controls B or B controls A. But this “control” relation can be cyclic, so it could happen that A controls B and B controls C and C controls A.

We can send in agents to infiltrate any particular cell, which gives us control over that cell and the cells that it controls, but not any other cells. So in the example above, infiltrating A would give us control over A and B, but not C.

For a successful infiltration of the ACM, we must obtain control over all of its cells, otherwise the cells that are out of our control will discover us and start causing some of their trademark chaos and mischief. As you know, we’re on a tight spending leash from higher authority these days, so we need to execute this mission as efficiently as possible. Your mission is to figure out the minimum number of cells we need to infiltrate in order to succeed.

This mission briefing will self-destruct in five hours. Good luck!

Input

The first line of a test case contains the number nn of cells the ACM has (1n751\le n\le 75). Each of the next nn lines contains a binary string of length nn where the ithi^{th} character of the jthj^{th} line is 11 if cell jj controls cell ii, and 00 otherwise (1i,jn1\le i, j \le n).

The ithi^{th} character of the ithi^{th} line is 00 and for i=ji = j, either the ithi^{th} character of the jthj^{th} line is 11 or the jthj^{th} character of the ithi^{th} line is 11, but not both.

Output

For each test case, display its case number followed by the minimum number mm of cells that must be infiltrated to obtain complete control of the ACM. Then display mm numbers c1,...,cmc_{1}, . . . , c_{m} in any order, indicating the list of cells to infiltrate (cells are numbered from 11 to nn). If more than one set of mm cells gives complete control, any one will be accepted.

ACM-ICPC World Finals 2012 Problem E: Infiltration

Sample Input

2
00
10
3
010
001
100
5
01000
00011
11001
10100
10010

Sample Output

Case 1: 1 2
Case 2: 2 1 2
Case 3: 2 2 3

ACM-ICPC World Finals 2012 Problem E: Infiltration

No official solution in the source collection.