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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring>
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>
#include <deque>

using namespace std;

#define mod 1000000007


int main()
{
	int n;
	cin >> n;
	vector<int> x, y;
	for(int i = 0; i < n; i++){
		if(i % 6 == 1){
			for(int j = 0; j < n; j += 2){
				x.push_back(i);
				y.push_back(j);
			}
		} else if(i % 6 == 4){
			for(int j = 1; j < n; j += 2){
				x.push_back(i);
				y.push_back(j);
			}
		} else if(i % 6 == 3 || i % 6 == 5){
			x.push_back(i);
			y.push_back(0);
		}
	}
	if((n - 1) % 6 == 0){
		for(int j = 1; j < n; j += 2){
			x.push_back(n - 1);
			y.push_back(j);
		}
	} else if((n - 1) % 6 == 3){
		for(int j = 2; j < n; j += 2){
			x.push_back(n - 1);
			y.push_back(j);
		}
	}
	cout << x.size() << endl;
	for(int i = 0; i < x.size(); i++){
		cout << x[i] << " " << y[i] << endl;
	}
	return 0;
}