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:
// D - Make Them Even
#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
#define rp(i,s,e) for(int i=(s);i<(e);++i)
#define rpz(i,e) rp(i,0,e)
using I4 = tuple<int,int,int,int>;
using VI4 = vector<I4>;

int main(){
	int H,W; cin>>H>>W;
	vvi A(H, vi(W)); rpz(i,H) rpz(j,W)cin>>A[i][j];

	VI4 YX;//[i]:{y,x,y',x'}
	int y1,x1,x2,y2;
	rpz(h, H) rpz(w, W){
		if(A[h][w] == 0) continue;
		if(A[h][w]%2 == 0) continue;
		y1 = y2 = h, x1 = x2 = w;
		if(w+1 == W) y2++; else x2++;
		y2 = min(H-1, y2), x2 = min(W-1, x2);
		if(y1==y2 && x1==x2) continue;
		A[y1][x1]--, A[y2][x2]++;
		YX.emplace_back(y1,x1,y2,x2);
	}

	cout<< YX.size() <<endl;
	for(auto t:YX){
		tie(y1,x1,y2,x2) = t;
		printf("%d %d %d %d\n",y1+1,x1+1,y2+1,x2+1);
	}
}