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 <string>
#include <vector>
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<stack>
#include<queue>
#include<cmath>
#include<algorithm>
#include<functional>
#include<list>
#include<deque>
#include<bitset>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<cstring>
#include<sstream>
#include<complex>
#include<iomanip>
#include<numeric>
#include<cassert>
#define X first
#define Y second
#define pb push_back
#define rep(X,Y) for (int (X) = 0;(X) < (Y);++(X))
#define reps(X,S,Y) for (int (X) = S;(X) < (Y);++(X))
#define rrep(X,Y) for (int (X) = (Y)-1;(X) >=0;--(X))
#define rreps(X,S,Y) for (int (X) = (Y)-1;(X) >= (S);--(X))
#define repe(X,Y) for ((X) = 0;(X) < (Y);++(X))
#define peat(X,Y) for (;(X) < (Y);++(X))
#define all(X) (X).begin(),(X).end()
#define rall(X) (X).rbegin(),(X).rend()
#define eb emplace_back
#define UNIQUE(X) (X).erase(unique(all(X)),(X).end())
#define Endl endl

using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
template<class T> using vv=vector<vector<T>>;
template<class T> ostream& operator<<(ostream &os, const vector<T> &t) {
os<<"{"; rep(i,t.size()) {os<<t[i]<<",";} os<<"}"<<endl; return os;}
template<class S, class T> ostream& operator<<(ostream &os, const pair<S,T> &t) { return os<<"("<<t.first<<","<<t.second<<")";}
template<class T> inline bool MX(T &l,const T &r){return l<r?l=r,1:0;}
template<class T> inline bool MN(T &l,const T &r){return l>r?l=r,1:0;}
#define out(args...){vector<string> a_r_g_s=s_p_l_i_t(#args, ','); e_r_r(a_r_g_s.begin(), args); }
vector<string> s_p_l_i_t(const string &s, char c){vector<string> v;int d=0,f=0;string t;for(char c:s){if(!d&&c==',')v.pb(t),t="";else t+=c;if(c=='\"'||c=='\'')f^=1;if(!f&&c=='(')++d;if(!f&&c==')')--d;}v.pb(t);return move(v);}
void e_r_r(vector<string>::iterator it) {}
template<typename T, typename... Args> void e_r_r(vector<string>::iterator it, T a, Args... args){ if(*it==" 1"||*it=="1") cerr<<endl; else cerr << it -> substr((*it)[0] == ' ', it -> length()) << " = " << a << ", "; e_r_r(++it, args...);}
const ll MOD=1e9+7;

int main(){
  ios_base::sync_with_stdio(false);
  cout<<fixed<<setprecision(0);
  int n;
  cin>>n;
  vector<pii> re;
  rep(i,n)if(i%3==1)rep(j,n)if((i+j)%2) re.eb(i,j);
  rep(i,n)if(i%2) re.eb(i,0);
  if(n%3==1)rep(j,n)if((n-1+j)%2) re.eb(n-1,j);
  sort(all(re)); UNIQUE(re);
  cout<<re.size()<<endl;
  for(pii p:re) cout<<p.X<<" "<<p.Y<<endl;
  return 0;
}