CodeForces #419（div2）C. Karen and Game

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 number 0.

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 to g_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 and m (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 the i-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 the x-th row".
• col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column".

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

Examples

input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1

output

4
row 1
row 1
col 4
row 3

input

3 3
0 0 0
0 1 0
0 0 0

output

-1

input

3 3
1 1 1
1 1 1
1 1 1

output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 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.

题意：给你一个n行m列的二维数组，你可以对任意行或列执行一个删除操作，即使该行或列上所有数字-1，如果能使数组全为0，则输出最小操作数以及步骤，不能则输出-1.

通过思考，有以下规则：

1.一行中最小的数字，就是能对该行执行的最大操作次数，如2 3 4 3，最多操作2次后变为0 1 2 1，此后对该行无法继续操作，对于列也是一样。

2.若对所有行执行操作，等价于对所有列执行操作

3.把对所有行或所有列的同步操作优先进行，可以达到最简化的效果

最简化操作以行列数量为准，如n=2，m=4,数组如下

k+0,k+0,k+1,k+0

k+2,k+2,k+3,k+2

若要先消除k，操作行，使数组总和-m，操作列，使数组总和-n，因此行数少，则优先列，列数少，则操作行

 4.对于最简化后的数组，其任意数字等于其最大行操作数与最大列操作数之和。如上述例子

0 0 1 0

2 2 3 2

第二行为2，第三列为1，其余全为0

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<set>
using namespace std;
const int MAX=1e5+10;
#define INF 0x7fffffff
#define ll long long
#define FOR(i,n) for(i=1;i<=n;i++)

int main()
{
    int n,m,i,j,k,ans=0,a[105][105],row[105],col[105],mark1;
    cin>>n>>m;
    for(i=1;i<=n;i++)
    {
        row[i]=INF;
        for(j=1;j<=m;j++)
        {
            cin>>a[i][j];
            row[i]=min(row[i],a[i][j]);
        }
    }
    for(i=1;i<=m;i++)
    {
        col[i]=INF;
        for(j=1;j<=n;j++)
        {
            col[i]=min(col[i],a[j][i]);
        }
    }
    if(n>m)
    {
        mark1=INF;
        for(i=1;i<=m;i++)
        {
            ans+=col[i];
            mark1=min(mark1,col[i]);
        }
        for(i=1;i<=n;i++)
        {
            row[i]-=mark1;
            ans+=row[i];
        }
    }
    else
    {
        mark1=INF;
        for(i=1;i<=n;i++)
        {
            ans+=row[i];
            mark1=min(mark1,row[i]);
        }
        for(i=1;i<=m;i++)
        {
            col[i]-=mark1;
            ans+=col[i];
        }
    }
    for(i=1;i<=n;i++)
    {
        for(j=1;j<=m;j++)
        {
            if(a[i][j]!=(col[j]+row[i]))
            {
                cout<<"-1"<<endl;
                return 0;
            }
        }
    }
    if(n>m)
    {
        cout<<ans<<endl;
        for(i=1;i<=m;i++)
            for(j=1;j<=col[i];j++)
            cout<<"col "<<i<<endl;
        for(i=1;i<=n;i++)
            for(j=1;j<=row[i];j++)
            cout<<"row "<<i<<endl;    
    }
    else
    {
        cout<<ans<<endl;
        for(i=1;i<=n;i++)
            for(j=1;j<=row[i];j++)
            cout<<"row "<<i<<endl;
        for(i=1;i<=m;i++)
            for(j=1;j<=col[i];j++)
            cout<<"col "<<i<<endl;
    }
}