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;
int main() {
    int n; cin >> n;
    vector<pair<int, int>> x;
    for (int i = 0; i < n; i++) {
        if (i % 6 == 1) {
            for (int j = 0; j < n; j++) {
                if (j % 2 == 0) {
                    x.push_back({i, j});
                }
            }
        } else if (i % 6 == 4) {
            for (int j = 0; j < n; j++) {
                if (j % 2 == 1) {
                    x.push_back({i, j});
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        if ((n-1 + i) % 2 == 1) {
            x.push_back({i, n-1});
            x.push_back({n-1, i});
        }
        if ((0 + i) % 2 == 1) {
            x.push_back({i, 0});
            x.push_back({0, i});
        }
    }
    sort(x.begin(), x.end());
    x.erase(unique(x.begin(), x.end()), x.end());
    cout << x.size() << endl;
    for (int i = 0; i < x.size(); i++) {
        cout << x[i].first << " " << x[i].second << endl;
    }
    return 0;
}