[CodeForces] CF226D The table

Harry Potter has a difficult homework. Given a rectangular table, consisting of n × m cells. Each cell of the table contains the integer. Harry knows how to use two spells: the first spell change the sign of the integers in the selected row, the second — in the selected column. Harry's task is to make non-negative the sum of the numbers in each row and each column using these spells.

Alone, the boy can not cope. Help the young magician!

Input

The first line contains two integers n and m (1 ≤ n,  m ≤ 100) — the number of rows and the number of columns.

Next n lines follow, each contains m integers: j-th integer in the i-th line is a_i, j (|a_i, j| ≤ 100), the number in the i-th row and j-th column of the table.

The rows of the table numbered from 1 to n. The columns of the table numbered from 1 to m.

Output

In the first line print the number a — the number of required applications of the first spell. Next print a space-separated integers — the row numbers, you want to apply a spell. These row numbers must be distinct!

In the second line print the number b — the number of required applications of the second spell. Next print b space-separated integers — the column numbers, you want to apply a spell. These column numbers must be distinct!

If there are several solutions are allowed to print any of them.

Examples

Input

Copy

4 1
-1
-1
-1
-1

Output

Copy

4 1 2 3 4 
0 

Input

Copy

2 4
-1 -1 -1 2
1 1 1 1

Output

Copy

1 1 
1 4 

题目解析

输出有点玄学。

大概意思就是先输出行、列最小交换几次，然后输出交换了哪一行

我的做法是贪心，反复judge每一行、列的和，发现负的就翻转次数++，不停地翻转小于0的行列，总会让每行每列都变得大于0的。

最后，因为翻转两次=不翻转，所以最后要输出次数&1。挺好玩的，神题。

Code

#include<iostream>
#include<cstdio>
using namespace std;

int n,m;
int a[105][105];
int sumx[105],sumy[105];
int flagx[105],flagy[105];
int ans1,ans2;

inline int judge_x() {
    for(int i = 1;i <= n;i++) {
        if(sumx[i] < 0) return i;
    }
    return 0;
}

inline int judge_y() {
    for(int i = 1;i <= m;i++) {
        if(sumy[i] < 0) return i;
    }
    return 0;
}

int main() {
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++) {
        for(int j = 1;j <= m;j++) {
            scanf("%d",&a[i][j]);
            sumx[i] += a[i][j];
            sumy[j] += a[i][j];
        }
    }
    while(1) {
        int tmpx = judge_x();
        int tmpy = judge_y();
        if(!tmpx && !tmpy) break;
        if(tmpx) {
            flagx[tmpx]++;
            sumx[tmpx] = 0;
            for(int i = 1;i <= m;i++) {
                int k = -a[tmpx][i];
                sumy[i] = sumy[i] - a[tmpx][i] + k;
                a[tmpx][i] = k;
                sumx[tmpx] += k;
            }
        } else {
            flagy[tmpy]++;
            sumy[tmpy] = 0;
            for(int i = 1;i <= n;i++) {
                int k = -a[i][tmpy];
                sumx[i] = sumx[i] - a[i][tmpy] + k;
                a[i][tmpy] = k;
                sumy[tmpy] += k;
            }
        }
    }
    for(int i = 1;i <= n;i++) if(flagx[i] & 1) ans1++;
    for(int i = 1;i <= m;i++) if(flagy[i] & 1) ans2++;
    printf("%d ",ans1);
    for(int i = 1;i <= n;i++) {
        if(flagx[i] & 1) printf("%d ",i);
    }
    printf("
%d ",ans2);
    for(int i = 1;i <= m;i++) {
        if(flagy[i] & 1) printf("%d ",i);
    }
    printf("
");
    return 0;
}