Toy Cars （周赛2）

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 matrixА: there is a number on the intersection of theі-th row and j-th column that describes the result of the collision of theі-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 A_ij = 1, then A_ji = 2, if A_ij = 3, then A_ji = 3, and if A_ij = 0, then A_ji = 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 0
0 -1 1
0 2 -1

Output

2
1 3 

Input

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

Output

0

大体题意就是有N辆车，组成一个N*N的矩阵，行和列的车进行碰撞，矩阵的主对角线只能是-1，如果为1,则行方向的车被撞翻，为2列方向的车被撞翻，为3行列都没撞翻，0是行列的车都被撞翻，第一行输出没有损坏的车有多少，第二行输出车对应的编号。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


int map[100][100],a[100],b[100];

int main()
{
    int n,i,j;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));

        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
            {
                scanf("%d",&map[i][j]);

                if (map[i][j]==1)
                    a[i]=1;
                else if (map[i][j]==2)
                    a[j]=1;
                else if (map[i][j]==3)
                {
                    a[i]=1;
                    a[j]=1;
                }
            }

            int sum=0;

            for(i=0;i<n;i++)
            {
                if(a[i]==0)
                {
                    b[sum++]=i+1;
                }

            }

            printf("%d\n",sum);

            for(i=0;i<sum;i++)
            {
                if(i!=sum-1)
                    printf("%d ",b[i]);
                else
                    printf("%d\n",b[i]);
            }

    }
    return 0;
}