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;
using Int = long long;
//INSERT ABOVE HERE
int s[1010][1010];
signed main(){
  int n;
  cin>>n;
  memset(s,0,sizeof(s));
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if((i+j)%2==0) s[i][j]=1;
  for(int i=0;i<n;i+=3)
    for(int j=0;j<n;j++)
      s[i][j]=1;
  for(int j=0;j<n;j++)
    s[n-1][j]=1;
  for(int i=0;i<n;i++)
    s[i][0]=1;

  using P = pair<int, int>;
  vector<P> ans;
  for(int i=0;i<n;i++)
    for(int j=0;j<n;j++)
      if(((i+j)&1)&&s[i][j]) ans.emplace_back(i,j);

  cout<<ans.size()<<endl;
  for(auto p:ans) cout<<p.first<<" "<<p.second<<endl;
  return 0;
}