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;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

#define fi first
#define se second
#define repl(i,a,b) for(ll i=(ll)(a);i<(ll)(b);i++)
#define rep(i,n) repl(i,0,n)
#define each(itr,v) for(auto itr:v)
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define dbg(x) cout<<#x"="<<x<<endl
#define mmax(x,y) (x>y?x:y)
#define mmin(x,y) (x<y?x:y)
#define maxch(x,y) x=mmax(x,y)
#define minch(x,y) x=mmin(x,y)
#define uni(x) x.erase(unique(all(x)),x.end())
#define exist(x,y) (find(all(x),y)!=x.end())
#define bcnt __builtin_popcount

#define INF INT_MAX/3

struct UnionFind{
  vector<int> v;
  UnionFind(int n) : v(n, -1) {}
  void init(){ for(int i = 0;i < v.size();i++)v[i]=-1; }
  int find(int x) { return v[x] < 0 ? x : v[x] = find(v[x]); }
  bool unite(int x, int y) {
    x = find(x); y = find(y);
    if (x == y) return false;
    if (-v[x] < -v[y]) swap(x, y);
    v[x] += v[y]; v[y] = x;
    return true;
  }
  bool root(int x) { return v[x] < 0; }
  bool same(int x, int y) { return find(x) == find(y); }
  int size(int x) { return -v[find(x)]; }
};

ll n;
ll dd[]={-1,0,1,0,-1};
vector<P> res;
ll tot;

bool valid(ll i,ll j){
  return i>=0&&i<n&&j>=0&&j<n;
}

int main(){
  cin>>n;
  rep(i,n)rep(j,n){
    if((i+j)%2==0)continue;
    tot++;
  }
  UnionFind uf(n*n);
  rep(hoge,7){
    rep(i,n)rep(j,n){
      if((i+j)%2==0)continue;
      bool ok=true;
      ll cnt=0;
      rep(d1,4)repl(d2,d1+1,4){
        ll ni1=i+dd[d1],nj1=j+dd[d1+1];
        ll ni2=i+dd[d2],nj2=j+dd[d2+1];
        if(!valid(ni1,nj1)||!valid(ni2,nj2)||(ni1==ni2&&nj1==nj2))continue;
        if(!uf.same(ni1*n+nj1,ni2*n+nj2))cnt++;
      }
      if(cnt>=7-hoge){
        res.push_back(P(i,j));
        rep(d1,4)rep(d2,4){
          ll ni1=i+dd[d1],nj1=j+dd[d1+1];
          ll ni2=i+dd[d2],nj2=j+dd[d2+1];
          if(!valid(ni1,nj1)||!valid(ni2,nj2))continue;
          uf.unite(ni1*n+nj1,ni2*n+nj2);
        }
      }
    }
  }
  printf("%lld\n", (ll)res.size());
  rep(i,res.size())printf("%lld %lld\n", res[i].se,res[i].fi);
  return 0;
}