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 <iostream>
#include <string>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <iomanip>


using namespace std;

#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define REPO(i, n) for(int i = 1;i <= n;i++)
#define ll long long
#define INF 1999999999
#define MINF -1999999999
#define INF64 1999999999999999999
#define ALL(n) n.begin(),n.end()
#define MOD 1000000007


ll n;
vector<pair<ll, ll>> s;
int main() {
	cin >> n;
	REP(i, n) {
		if (i % 3 == 0 or i == 0 or i == n - 1) {
			REP(j, n)if ((i + j) % 2 == 1)s.push_back(make_pair(j, i));
		}
		else if (i % 2 == 1) s.push_back(make_pair(0, i));
	}
	cout << s.size() << endl;
	REP(i, s.size()) cout << s[i].first << " " << s[i].second << endl;
}