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[505][505];
vector<int>ans;
int main()
{
    int h,w;
    scanf("%d%d",&h,&w);
    for(int i=1;i<=h;i++)
    	for(int j=1;j<=w;j++)
    		scanf("%d",&a[i][j]);
    for(int i=1;i<h;i++)
    {
    	for(int j=1;j<=w;j++)
    	{
    		if(a[i][j]%2)
    		{
    			ans.push_back(i);
    			ans.push_back(j);
    			ans.push_back(i+1);
    			ans.push_back(j);
    			a[i+1][j]++;
			}
		}
	}
	for(int i=1;i<w;i++)
	{
		if(a[h][i]%2)
		{
			ans.push_back(h);
			ans.push_back(i);
			ans.push_back(h);
			ans.push_back(i+1);
			a[h][i+1]++;
		}
	}
	printf("%d\n",ans.size()/4);
	for(int i=0;i<ans.size();i+=4)
		printf("%d %d %d %d\n",ans[i],ans[i+1],ans[i+2],ans[i+3]);
	return 0;
}