CodeForces

Problem Description

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 gi, 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 gi, j (0 ≤ gi, 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.

解题思路：贪心模拟，行操作完再进行列操作。注意：题目要求用最少的步数，因此当n>m时，应先对列进行贪心减操作，再对行进行操作；否则先对行进行贪心减操作，最后如果减不完，则直接输出-1.

AC代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=105;
const int inf=500;
int n,m,cnt1,cnt2,ans,sum,mp[maxn][maxn],min_row[maxn],min_col[maxn];
pair<int,int> paii_row[maxn],paii_col[maxn];
int main(){
    while(cin>>n>>m){
        ans=sum=cnt1=cnt2=0;
        for(int i=1;i<=n;++i)min_row[i]=inf;
        for(int j=1;j<=m;++j)min_col[j]=inf;
        for(int i=1;i<=n;++i){
            for(int j=1;j<=m;++j){
                cin>>mp[i][j],sum+=mp[i][j];
                min_row[i]=min(min_row[i],mp[i][j]);
                min_col[j]=min(min_col[j],mp[i][j]);
            }
        }
        if(n>m){///如果行大于列，则从列开始操作
            for(int j=1;j<=m;++j){
                for(int i=1;i<=n;++i){
                    mp[i][j]-=min_col[j];
                    min_row[i]=min(min_row[i],mp[i][j]);
                }
                if(min_col[j])sum-=min_col[j]*n,ans+=(paii_col[cnt2].first=min_col[j]),paii_col[cnt2++].second=j,min_col[j]=0;
            }
            for(int i=1;i<=n;++i){///从行开始操作
                for(int j=1;j<=m;++j){
                    mp[i][j]-=min_row[i];
                    min_col[j]=min(min_col[j],mp[i][j]);
                }
                if(min_row[i])sum-=min_row[i]*m,ans+=(paii_row[cnt1].first=min_row[i]),paii_row[cnt1++].second=i,min_row[i]=0;
            }
        }
        else{
            for(int i=1;i<=n;++i){///否则先从行开始操作
                for(int j=1;j<=m;++j){
                    mp[i][j]-=min_row[i];
                    min_col[j]=min(min_col[j],mp[i][j]);
                }
                if(min_row[i])sum-=min_row[i]*m,ans+=(paii_row[cnt1].first=min_row[i]),paii_row[cnt1++].second=i,min_row[i]=0;
            }
            for(int j=1;j<=m;++j){
                for(int i=1;i<=n;++i){
                    mp[i][j]-=min_col[j];
                    min_row[i]=min(min_row[i],mp[i][j]);
                }
                if(min_col[j])sum-=min_col[j]*n,ans+=(paii_col[cnt2].first=min_col[j]),paii_col[cnt2++].second=j,min_col[j]=0;
            }
        }
        if(sum)cout<<-1<<endl;
        else{
            cout<<ans<<endl;
            for(int i=0;i<cnt1;++i)
                for(int j=1;j<=paii_row[i].first;++j)
                    cout<<"row "<<paii_row[i].second<<endl;
            for(int i=0;i<cnt2;++i)
                for(int j=1;j<=paii_col[i].first;++j)
                    cout<<"col "<<paii_col[i].second<<endl;
        }
    }
    return 0;
}