Problem: There is a grid of square cells with H horizontal rows and W vertical columns. The cell at the i-th row and the j-th column will be denoted as Cell (i, j).

In Cell (i, j), a_{ij} coins are placed.

You can perform the following operation any number of times:

Operation: Choose a cell that was not chosen before and contains one or more coins, then move one of those coins to a vertically or horizontally adjacent cell.

Maximize the number of cells containing an even number of coins.

Constraints

* All values in input are integers.
* 1 \leq H, W \leq 500
* 0 \leq a_{ij} \leq 9

Input

Input is given from Standard Input in the following format:


H W
a_{11} a_{12} ... a_{1W}
a_{21} a_{22} ... a_{2W}
:
a_{H1} a_{H2} ... a_{HW}


Output

Print a sequence of operations that maximizes the number of cells containing an even number of coins, in the following format:


N
y_1 x_1 y_1' x_1'
y_2 x_2 y_2' x_2'
:
y_N x_N y_N' x_N'


That is, in the first line, print an integer N between 0 and H \times W (inclusive), representing the number of operations.

In the (i+1)-th line (1 \leq i \leq N), print four integers y_i, x_i, y_i' and x_i' (1 \leq y_i, y_i' \leq H and 1 \leq x_i, x_i' \leq W), representing the i-th operation. These four integers represents the operation of moving one of the coins placed in Cell (y_i, x_i) to a vertically or horizontally adjacent cell, (y_i', x_i').

Note that if the specified operation violates the specification in the problem statement or the output format is invalid, it will result in Wrong Answer.

Examples

Input

2 3
1 2 3
0 1 1


Output

3
2 2 2 3
1 1 1 2
1 3 1 2


Input

3 2
1 0
2 1
1 0


Output

3
1 1 1 2
1 2 2 2
3 1 3 2


Input

1 5
9 9 9 9 9


Output

2
1 1 1 2
1 3 1 4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int a[510][510];
struct move{
	int srx,sry,tx,ty;
}mv[250010];
int main(){
	//freopen("matrix.in","r",stdin);
	//freopen("matrix.out","w",stdout);
	int n,m,k=0,i,j,count=0;
	cin>>n>>m;
	for(i=1;i<=n;i++)
	  for(j=1;j<=m;j++)
	    cin>>a[i][j];
	memset(mv,0,sizeof(mv));
	for(i=1;i<=n;i++)
	  for(j=1;j<=m-1;j++)
	    if(a[i][j] % 2==1){
	    	a[i][j]--;
	    	a[i][j+1]++;
	    	k++;
	    	mv[k].srx=i;mv[k].sry=j;mv[k].tx=i;mv[k].ty=j+1;
	    	count++;
	    }
	for(i=1;i<=n-1;i++)
	  if(a[i][m]%2==1){
	  	a[i][m]--;
	  	a[i+1][m]++;
	  	k++;
	  	mv[k].srx=i,mv[k].sry=m,mv[k].tx =i+1,mv[k].ty=m;
	  	count++;
	  }
	cout<<count<<endl;
	for(i=1;i<=k;i++)
	  cout<<mv[i].srx<<" "<<mv[i].sry<<" "<<mv[i].tx <<" "<<mv[i].ty <<endl;
	return 0;
}