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(){
  vector<vector<int>> f;
  int N, K = 0;

  cin>>N;

  f.resize(N, vector<int>(N));

  for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      f[i][j] = (i+j)%2;
    }
  }

  for(int i = 0; i < N; i++){
    if(f[0][i] == 1) f[0][i] = 2, K++;
  }

  for(int i = 1; i < N; i+=3){
    for(int j = 0; j < N; j++){
      // cout<<j<<" "<<i<<endl;
      if(f[j][i] == 1) f[j][i] = 2, K++;
    }
  }


  for(int j = 1; j < N; j++){
      if(f[j][N-1] == 1) f[j][N-1] = 2, K++;
  }
  /*
  for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      cout<<f[i][j];
    }cout<<endl;
  }
  */


  cout<<K<<endl;

   for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      if(f[i][j] == 2) {
         cout<<i<<" "<<j<<endl;

      }
    }
  }

  return 0;
}