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>
#define pir pair<int,int>
#define pb push_back
#define fi first
#define se second
using namespace std;

const int N=510;
int h,w,num,foo[N];
vector<pair<pir,pir>> ans;

int main() {
	scanf("%d%d",&h,&w);
	for(int i=1,j=0;i<=h;i++) {
		scanf("%d",&num);
		int tmp=num&1;
		for(int j=2;j<=w;j++) {
			scanf("%d",&num);
			if(tmp) ans.pb({{i,j-1},{i,j}});
			if(num&1) tmp^=1;
		}
		if(j&1) ans.pb({{i-1,w},{i,w}});
		if(tmp&1) j^=1;
	}
	printf("%d\n",ans.size());
	for(auto p:ans) {
		printf("%d %d %d %d\n",p.fi.fi,p.fi.se,p.se.fi,p.se.se);
	}
	return 0;
}