Bob is playing a game named "Walk on Matrix".

In this game, player is given an n × m matrix A=(a_{i,j}), i.e. the element in the i-th row in the j-th column is a_{i,j}. Initially, player is located at position (1,1) with score a_{1,1}. 

To reach the goal, position (n,m), player can move right or down, i.e. move from (x,y) to (x,y+1) or (x+1,y), as long as player is still on the matrix.

However, each move changes player's score to the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the current score and the value at the position he moves to.

Bob can't wait to find out the maximum score he can get using the tool he recently learnt — dynamic programming. Here is his algorithm for this problem. 

<image>

However, he suddenly realize that the algorithm above fails to output the maximum score for some matrix A. Thus, for any given non-negative integer k, he wants to find out an n × m matrix A=(a_{i,j}) such that 

  * 1 ≤ n,m ≤ 500 (as Bob hates large matrix); 
  * 0 ≤ a_{i,j} ≤ 3 ⋅ 10^5 for all 1 ≤ i≤ n,1 ≤ j≤ m (as Bob hates large numbers); 
  * the difference between the maximum score he can get and the output of his algorithm is exactly k. 



It can be shown that for any given integer k such that 0 ≤ k ≤ 10^5, there exists a matrix satisfying the above constraints.

Please help him with it!

Input

The only line of the input contains one single integer k (0 ≤ k ≤ 10^5).

Output

Output two integers n, m (1 ≤ n,m ≤ 500) in the first line, representing the size of the matrix. 

Then output n lines with m integers in each line, a_{i,j} in the (i+1)-th row, j-th column.

Examples

Input


0


Output


1 1
300000

Input


1


Output


3 4
7 3 3 1
4 8 3 6
7 7 7 3

Note

In the first example, the maximum score Bob can achieve is 300000, while the output of his algorithm is 300000.

In the second example, the maximum score Bob can achieve is 7\&3\&3\&3\&7\&3=3, while the output of his algorithm is 2.
"""
    Author - Satwik Tiwari .
    2nd NOV , 2020  - Monday
"""

#===============================================================================================
#importing some useful libraries.


from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from functools import cmp_to_key

# from itertools import *
from heapq import *
from math import gcd, factorial,floor,ceil,sqrt

from copy import deepcopy
from collections import deque


from bisect import bisect_left as bl
from bisect import bisect_right as br
from bisect import bisect

#==============================================================================================
#fast I/O region
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")

#===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
  def wrapped_func(*args, **kwargs):
    if stack: return f(*args, **kwargs)
    to = f(*args, **kwargs)
    while True:
      if type(to) is GeneratorType:
        stack.append(to)
        to = next(to)
        continue
      stack.pop()
      if not stack: break
      to = stack[-1].send(to)
    return to
  return wrapped_func
#### END ITERATE RECURSION ####

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

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 pp in range(t):
        solve(pp)
def printlist(a) :
    for p in range(0,len(a)):
        out(str(a[p]) + ' ')
def google(p):
    print('Case #'+str(p)+': ',end='')
def lcm(a,b): return (a*b)//gcd(a,b)
def power(x, y, p) :
    y%=(p-1)  #not so sure about this. used when y>p-1. if p is prime.
    res = 1     # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0) :
        return 0
    while (y > 0) :
        if ((y & 1) == 1) : # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1      # y = y/2
        x = (x * x) % p
    return res
def ncr(n,r): return factorial(n) // (factorial(r) * factorial(max(n - r, 1)))
def isPrime(n) :
    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
inf = pow(10,20)
mod = 10**9+7
#===============================================================================================
# code here ;))

def solve(case):
    k = int(inp())
    temp = int('1'*(len(bin(100000))-1),2)
    val = '1' + '0'*(len(bin(k))-2)
    val = int(val,2)
    print(3,3)
    g = [[temp,k,0],[val,temp,0],[0,k,k]]

    for i in range(3):
        print(*g[i])

    #CHEKCING MY LOGIC IS CORRECT OR NOT xD
    # dp = deepcopy(g)
    # for i in range(3):
    #     print(dp[i])
    #
    # for i in range(3):
    #     for j in range(3):
    #         if(i == 0 and j == 0):
    #             continue
    #         if(0<=i-1<3 and 0<=j-1<3):
    #             if(i == 1 and j == 1):
    #                 print(g[i][j],dp[i-1][j],dp[i][j-1])
    #             dp[i][j] = max(g[i][j]&dp[i-1][j],g[i][j]&dp[i][j-1])
    #         else:
    #             if(0<=i-1<3):
    #                 dp[i][j] = dp[i][j] & dp[i-1][j]
    #             elif(0<=j-1<3):
    #                 dp[i][j] = dp[i][j] & dp[i][j-1]
    #
    # print('======')
    # for i in range(3):
    #     print(dp[i])
    #
    # print(k,dp[2][2])



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









