Karen and Game Codeforces

C. Karen and Game

time limit per test

2 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

On the way to school, Karen became fixated on the puzzle game on her phone!

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal tog_i, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n andm (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in thei-th of these rows contains g_i, j (0 ≤ g_i, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer-1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

• row x, (1 ≤ x ≤ n) describing a move of the form "choose thex-th row".
• col x, (1 ≤ x ≤ m) describing a move of the form "choose thex-th column".

If there are multiple optimal solutions, output any one of them.

Examples

Input

3 52 2 2 3 20 0 0 1 01 1 1 2 1

Output

4row 1row 1col 4row 3

Input

3 30 0 00 1 00 0 0

Output

-1

Input

3 31 1 11 1 11 1 1

Output

3row 1row 2row 3

Note

In the first test case, Karen has a grid with 3 rows and5 columns. She can perform the following 4 moves to beat the level:

In the second test case, Karen has a grid with 3 rows and3 columns. It is clear that it is impossible to beat the level; performing any move will create three1s on the grid, but it is required to only have one1 in the center.

In the third test case, Karen has a grid with 3 rows and3 columns. She can perform the following 3 moves to beat the level:

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

题目大意：

给你一个初始化都是0的矩阵，让你用最小的步数变成目标矩阵。

转换一下，就是把目标矩阵用最小的步骤数变为0矩阵。每次都只能变一行或者一列，并且选定的只能行或者列一次只能减去1.

如果我们从行开始删除，从1-n逐行删除最小的，再从列1-m删除，这样的话一定可以全部删除为。但是这样不一定删除的步数是最小的。

举个例子把

11111111

11111111

11111111

我先从n开始删除只需要删除3次

但是我从列开始删除呢？

我需要删除8次

所以我们选择从行开始删除，这时候行列的关系是n<m

也就是从小的开始删除

开始看着长，其实就是复制粘贴

下面AC代码：

#include<bits/stdc++.h>using namespace std;int save[101][101];int put1[10100000];int put2[10001000];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)                scanf("%d",&save[i][j]);        int c1=0,c2=0;        if(m>n)        {            for(int i=1; i<=n; i++)            {                int minn=0x3f3f3f3f;                for(int j=1; j<=m; j++)                {                    if(save[i][j]<minn)                        minn=save[i][j];                }                if(minn>0)                {                    for(int j=1; j<=m; j++)                    {                        save[i][j]-=minn;                    }                    while(minn--)                    {                        put1[c1++]=i;                    }                }            }            for(int i=1; i<=m; i++)            {                int minn=0x3f3f3f3f;                for(int j=1; j<=n; j++)                {                    if(save[j][i]<minn)                        minn=save[j][i];                }                if(minn>0)                {                    for(int j=1; j<=n; j++)                    {                        save[j][i]-=minn;                    }                    while(minn--)                    {                        put2[c2++]=i;                    }                }            }        }        else        {            for(int i=1; i<=m; i++)            {                int minn=0x3f3f3f3f;                for(int j=1; j<=n; j++)                {                    if(save[j][i]<minn)                        minn=save[j][i];                }                if(minn>0)                {                    for(int j=1; j<=n; j++)                    {                        save[j][i]-=minn;                    }                    while(minn--)                    {                        put2[c2++]=i;                    }                }            }            for(int i=1; i<=n; i++)            {                int minn=0x3f3f3f3f;                for(int j=1; j<=m; j++)                {                    if(save[i][j]<minn)                        minn=save[i][j];                }                if(minn>0)                {                    for(int j=1; j<=m; j++)                    {                        save[i][j]-=minn;                    }                    while(minn--)                    {                        put1[c1++]=i;                    }                }            }        }        int flag=1;        for(int i=1; i<=n; i++)            for(int j=1; j<=m; j++)            {                if(save[i][j])                    flag=0;            }        if(flag)        {            printf("%d\n",c1+c2);            for(int i=0; i<c1; i++)            {                printf("row %d\n",put1[i]);            }            for(int i=0; i<c2; i++)            {                printf("col %d\n",put2[i]);            }        }        else            printf("-1\n");    }}