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;

#define se second
#define fi first
#define mk(a,b) make_pair(a,b)

int N;
set<pair<int,int> > st;
vector<pair<int,int> > ans;

int main()
{
    scanf("%d",&N);

    for(int i = 1; i < N; i += 3)
    {
        for(int j = 1 - (i % 2); j < N; j += 2)
        {
            st.insert(mk(j,i));
            ans.push_back(mk(j,i));
        }
    }

    for(int i = 0; i < N; i++)
    {
        if(i % 2 == 0)continue;
        if(st.count(mk(0,i)))continue;
        ans.push_back(mk(0,i));
        st.insert(mk(0,i));
    }
    for(int i = 0; i < N; i++)
    {
        if((i + N - 1) % 2 == 0)continue;
        if(st.count(mk(i,N - 1)))continue;
        ans.push_back(mk(i,N - 1));
        st.insert(mk(i,N - 1));
    }

    printf("%d\n",ans.size());
    for(int i = 0; i < ans.size(); i++)
    {
        printf("%d %d\n",ans[i].fi,ans[i].se);
    }
    return 0;
}