Problem: Leo Jr. draws pictures in his notebook with checkered sheets (that is, each sheet has a regular square grid printed on it). We can assume that the sheets are infinitely large in any direction.

To draw a picture, Leo Jr. colors some of the cells on a sheet gray. He considers the resulting picture beautiful if the following conditions are satisfied:

  * The picture is connected, that is, it is possible to get from any gray cell to any other by following a chain of gray cells, with each pair of adjacent cells in the path being neighbours (that is, sharing a side).
  * Each gray cell has an even number of gray neighbours.
  * There are exactly n gray cells with all gray neighbours. The number of other gray cells can be arbitrary (but reasonable, so that they can all be listed).



Leo Jr. is now struggling to draw a beautiful picture with a particular choice of n. Help him, and provide any example of a beautiful picture.

To output cell coordinates in your answer, assume that the sheet is provided with a Cartesian coordinate system such that one of the cells is chosen to be the origin (0, 0), axes 0x and 0y are orthogonal and parallel to grid lines, and a unit step along any axis in any direction takes you to a neighbouring cell.

Input

The only line contains a single integer n (1 ≤ n ≤ 500) — the number of gray cells with all gray neighbours in a beautiful picture.

Output

In the first line, print a single integer k — the number of gray cells in your picture. For technical reasons, k should not exceed 5 ⋅ 10^5.

Each of the following k lines should contain two integers — coordinates of a gray cell in your picture. All listed cells should be distinct, and the picture should satisdfy all the properties listed above. All coordinates should not exceed 10^9 by absolute value.

One can show that there exists an answer satisfying all requirements with a small enough k.

Example

Input


4


Output


12
1 0
2 0
0 1
1 1
2 1
3 1
0 2
1 2
2 2
3 2
1 3
2 3

Note

The answer for the sample is pictured below:

<image>

Solution (in PYTHON):

Solution:
"""
    Satwik_Tiwari ;) .
    16 june , 2020  - Tuesday
"""

#===============================================================================================
#importing some useful libraries.
from __future__ import division, print_function

from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase


import bisect
from heapq import *
from math import *
from collections import deque
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations as comb # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
#If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
#If the element is already present in the list,
# the right most position where element has to be inserted is returned

#==============================================================================================

BUFSIZE = 8192


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


def print(*args, **kwargs):
    """Prints the values to a stream, or to sys.stdout by default."""
    sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
    at_start = True
    for x in args:
        if not at_start:
            file.write(sep)
        file.write(str(x))
        at_start = False
    file.write(kwargs.pop("end", "\n"))
    if kwargs.pop("flush", False):
        file.flush()


if sys.version_info[0] < 3:
    sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
    sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

#===============================================================================================
#some shortcuts

mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") #for fast input
def out(var): sys.stdout.write(str(var))  #for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def graph(vertex): return [[] for i in range(0,vertex+1)]
def zerolist(n): return [0]*n
def nextline(): out("\n")  #as stdout.write always print sring.
def testcase(t):
    for p in range(t):
        solve()
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(a,b):
    ans = 1
    while(b>0):
        if(b%2==1):
            ans*=a
        a*=a
        b//=2
    return ans
def ncr(n,r): return factorial(n)//(factorial(r)*factorial(max(n-r,1)))
def isPrime(n) : # Check Prime Number or not
    if (n <= 1) : return False
    if (n <= 3) : return True
    if (n % 2 == 0 or n % 3 == 0) : return False
    i = 5
    while(i * i <= n) :
        if (n % i == 0 or n % (i + 2) == 0) :
            return False
        i = i + 6
    return True

#===============================================================================================
# code here ;))
def printa(a):
    s = ''
    for i in range(len(a)):
        s += a[i]
    print(s)

def solve():
    n = int(inp())
    coor = []
    for i in range(0,(2*n)+1):
        coor.append([i,2])
    for i in range(1,2*n,2):
        coor.append([i,1])
        coor.append([i,3])
    if(n%2 == 0):
        for i in range(3,(2*n)-1,4):
            coor.append([i,4])
            coor.append([i+1,4])
            coor.append([i+2,4])
            coor.append([i,0])
            coor.append([i+1,0])
            coor.append([i+2,0])
        coor.append([0,1])
        coor.append([2*n,1])
        coor.append([1,4])
        coor.append([(2*n)-1,4])
        coor.append([1,5])
        coor.append([(2*n)-1,5])
        for i in range(1,2*n):
            coor.append([i,6])
    else:
        for i in range(1,2*n-1,4):
            coor.append([i,4])
            coor.append([i+1,4])
            coor.append([i+2,4])
        coor.append([2*n,3])
        for i in range(3,2*n,4):
            coor.append([i,0])
            coor.append([i+1,0])
            coor.append([i+2,0])
        coor.append([0,1])

    out(len(coor))
    nextline()
    for i in range(0,len(coor)):
        out(str(coor[i][0])+' '+ str(coor[i][1]))
        nextline()

testcase(1)
# testcase(int(inp()))