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 "bits/stdc++.h"
using namespace std;

int main() {
	long long N;
	cin >> N;
	vector<pair<long long, long long> > ANS;
	for (int i = 1; i < N; i += 2) {
		ANS.push_back({ 0, i });
	}
	for (int i = 1; i < N; i += 3) {
		if (i % 2 == 0) {
			for (int j = 1; j < N; j += 2) ANS.push_back({ j, i });
		}
		else {
			for (int j = 2; j < N; j += 2) ANS.push_back({ j, i });
		}
	}
	if (N % 3 == 1) {
		if (N % 2 == 0) {
			for (int i = 2; i < N; i += 2) ANS.push_back({ i, N - 1 });
		}
		else {
			for (int i = 1; i < N; i += 2) ANS.push_back({ i, N - 1 });
		}
	}
	cout << ANS.size() << endl;
	for (pair<long long, long long > P : ANS) cout << P.first << " " << P.second << endl;
}