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;

struct Initializer {
  Initializer() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(15);
  }
} initializer;

int main() {
  int n;
  cin >> n;
  vector<pair<int, int>> v;
  for (int i = 1; i < n; i += 2) v.emplace_back(i, 0);
  for (int i = 1; i < n - 2; i += 6) {
    for (int j = 2; j < n; j += 2) v.emplace_back(i, j);
  }
  for (int i = 4; i < n - 2; i += 6) {
    for (int j = 1; j < n; j += 2) v.emplace_back(i, j);
  }
  
  for (int j = n % 2 + 1; j < n; j += 2) v.emplace_back(n - 2, j);
  cout << v.size() << endl;
  for (auto c : v) cout << c.first << " " << c.second << endl;
  /*
  vector<vector<int>> k(n, vector<int>(n));
  for (auto c : v) k[c.first][c.second] = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if ((i + j) % 2 == 0) k[i][j] = 1;
      cout << k[i][j];
    }
    cout << endl;
  }
  */
}