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;
typedef unsigned long long int ull;
typedef long long int ll;
typedef pair<ll,ll> pll;
typedef long double D;
//typedef complex<D> P;
#define F first
#define S second
const ll MOD=1000000007;
//const ll MOD=998244353;

template<typename T,typename U>istream & operator >> (istream &i,pair<T,U> &A){i>>A.F>>A.S; return i;}
template<typename T>istream & operator >> (istream &i,vector<T> &A){for(auto &I:A){i>>I;} return i;}
template<typename T,typename U>ostream & operator << (ostream &o,const pair<T,U> &A){o<<A.F<<" "<<A.S; return o;}
template<typename T>ostream & operator << (ostream &o,const vector<T> &A){int i=A.size(); for(auto &I:A){o<<I<<(--i?" ":"");} return o;}



int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll N;
  cin>>N;
  ll cnt=0;
  for(int i=1;i<N;i+=3){
    cnt+=(N+(i&1))/2;
  }
  if((N-1)%3==0){cnt+=(N+(N-1&1))/2;}
  for(int i=2;i<N;i+=3){cnt++;}
  cout<<cnt<<endl;
  for(int i=1;i<N;i+=3){
    for(int j=(i&1?0:1);j<N;j+=2){cout<<i<<" "<<j<<endl;}
  }
  if((N-1)%3==0){
    for(int j=(N-1&1?0:1);j<N;j+=2){cout<<N-1<<" "<<j<<endl;}
  }
  for(int i=2;i<N;i+=3){cout<<i<<" "<<(i&1?0:1)<<endl;}


  return 0;
}