A. Toy Cars

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Susie, thanks to her older brother, likes to play with cars. Today she decided to set up a tournament between them. The process of a tournament is described in the next paragraph.

There are n toy cars. Each pair collides. The result of a collision can be one of the following: no car turned over, one car turned over, both cars turned over. A car is good if it turned over in no collision. The results of the collisions are determined by an n × n matrix А: there is a number on the intersection of the і-th row and j-th column that describes the result of the collision of the і-th and the j-th car:

•  - 1: if this pair of cars never collided.  - 1 occurs only on the main diagonal of the matrix.
• 0: if no car turned over during the collision.
• 1: if only the i-th car turned over during the collision.
• 2: if only the j-th car turned over during the collision.
• 3: if both cars turned over during the collision.

Susie wants to find all the good cars. She quickly determined which cars are good. Can you cope with the task?

Input

The first line contains integer n (1 ≤ n ≤ 100) — the number of cars.

Each of the next n lines contains n space-separated integers that determine matrix A.

It is guaranteed that on the main diagonal there are  - 1, and  - 1 doesn't appear anywhere else in the matrix.

It is guaranteed that the input is correct, that is, if Aij = 1, then Aji = 2, if Aij = 3, then Aji = 3, and if Aij = 0, thenAji = 0.

Output

Print the number of good cars and in the next line print their space-separated indices in the increasing order.

Sample test(s)

input

3-1 0 00 -1 10 2 -1

output

21 3 

input

4-1 3 3 33 -1 3 33 3 -1 33 3 3 -1

output

0

解题说明：此题是一道模拟题，只需要判断出现1和出现3的情况，并把相应行所代表的玩具车进行标记，最后进行输出。

#include <stdio.h>int main(){int i,j,n,dl[101][101],kq[101],kq1[101],t=0;scanf("%d",&n);for(i=1;i<=n;i++){for(j=1;j<=n;j++){scanf("%d",&dl[i][j]);if(dl[i][j]==1){kq[i]=1;}if(dl[i][j]==3){kq[i]=1;}}}for(i=1;i<=n;i++){if(kq[i]==0){t++;kq1[t]=i;}}printf("%d\n",t);for(i=1;i<=t;i++){printf("%d ",kq1[i]);}return 0;}