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 <iostream>

using namespace std;
int a[505][505],vis[505][505],h,w,tot;
struct node{
	int x1,y1,x2,y2;
};
node step[250005];
int main(){
	ios::sync_with_stdio(false);
	cin >> h >> w;
	for(int i = 1;i <= h;++i)
		for(int j = 1;j <= w;++j)cin >> a[i][j];
	for(int i = 1;i <= h;++i)
		for(int j = 1;j <= w;++j)
		if(a[i][j] %2 == 0)continue;
		else if(i == h && j == w)continue;
		else {
			if(j == w){
				step[++tot] = {i,j,i + 1,j};
				a[i + 1][j]++;
			}
			else {
				step[++tot] = {i,j,i,j + 1};
				a[i][j + 1]++;
			}
		}
	cout << tot << endl;
	for(int i = 1;i <= tot;++i)
		cout << step[i].x1 << ' ' << step[i].y1 << ' ' << step[i].x2 << ' ' << step[i].y2 << endl;
	return 0;
}