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 <vector>
using namespace std;

typedef pair<int,int> P;
bool B[1000][1000];
vector<P> ANS;

int main(){
    int N;
    cin>>N;

    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if((i+j)%2==0){
                B[i][j]=true;
            }
        }
    }
    for(int i=0;i<N;i+=3){
        for(int j=0;j<N;j++){
            if(!B[i][j]){
                B[i][j]=true;
                ANS.emplace_back(i,j);
            }
        }
    }
    for(int i=0;i<N;i++){
        if(!B[N-1][i]){
            B[N-1][i]=true;
            ANS.emplace_back(N-1,i);
        }
    }
    for(int i=0;i<N;i++){
        if(!B[i][0]){
            B[i][0]=true;
            ANS.emplace_back(i,0);
        }
    }
    cout<<ANS.size()<<endl;
    for(auto ans:ANS){
        cout<<ans.first<<' '<<ans.second<<endl;
    }

    return 0;
}