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 <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <algorithm>
using namespace std;
 
#define REP(i,n) for(ll i=0; i<n; ++i)
#define FOR(i,a,b) for(ll i=a; i<=b; ++i)
#define FORR(i,a,b) for (ll i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()
 
typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef vector<VI> VVI;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;
 
int main() {
    int n;
    cin >> n;

    set<P> p;
    for (int j = 1; j < n; j += 2){
        p.insert(P(0, j));
    }
    FOR(i,1,n-2){
        if (i % 2){
            for (int j = 4; j < n; j += 6) p.insert(P(i, j));
        }else{
            for (int j = 1; j < n; j += 6) p.insert(P(i, j));
        }
    }

    REP(j,n){
        if ((n - 1 + j) % 2) p.insert(P(n-1, j));
    }
    REP(i,n){
        if (i % 2) p.insert(P(i, 0));
        if ((i + n - 1) % 2) p.insert(P(i, n-1));
    }

    cout << p.size() << endl;
    for (P pp : p){
        cout << pp.first << " " << pp.second << endl;
    }
 
    return 0;
}