Problem: We have an N×N checkerboard.

From the square at the upper left corner, a square that is i squares to the right and j squares below is denoted as (i, j). Particularly, the square at the upper left corner is denoted as (0, 0).

Each square (i, j) such that i+j is even, is colored black, and the other squares are colored white.

We will satisfy the following condition by painting some of the white squares:

* Any black square can be reached from square (0, 0) by repeatedly moving to a black square that shares a side with the current square.



Achieve the objective by painting at most 170000 squares black.

Constraints

* 1 \leq N \leq 1,000

Input

The input is given from Standard Input in the following format:


N


Output

Print the squares to paint in the following format:


K
x_1 y_1
x_2 y_2
:
x_K y_K


This means that a total of K squares are painted black, the i-th of which is (x_i, y_i).

Judging

The output is considered correct only if all of the following conditions are satisfied:

* 0 \leq K \leq 170000
* 0 \leq x_i, y_i \leq N-1
* For each i, x_i + y_i is odd.
* If i \neq j, then (x_i, y_i) \neq (x_j, y_j).
* The condition in the statement is satisfied by painting all specified squares.

Examples

Input

2


Output

1
1 0


Input

4


Output

3
0 1
2 1
2 3

Solution (in CPP):

Solution:
#include<stdio.h>
int x[170010],y[170010],M;
void put(int a,int b){
	M++;
	x[M]=a;
	y[M]=b;
}
int main(){
	int n,i,j;
	scanf("%d",&n);
	for(i=0;6*i+1<n;i++){
		if(6*i+1<n){
			for(j=0;j<n;j+=2)put(6*i+1,j);
		}
		if(6*i+3<n)put(6*i+3,0);
		if(6*i+4<n){
			for(j=1;j<n;j+=2)put(6*i+4,j);
		}
		if(6*i+5<n)put(6*i+5,0);
	}
	if(n%6==4){
		for(i=2;i<n;i+=2)put(n-1,i);
	}
	if(n%6==1){
		for(i=2;i<n;i+=2)put(n-2,i);
	}
	printf("%d\n",M);
	for(i=1;i<=M;i++)printf("%d %d\n",x[i],y[i]);
}