human
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. 


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.
#from
assistant
#Here is Python program to find out the maximum score he can get using the tool he recently learnt — dynamic programming.
 __future__ import print_function
mod=int(1e9+7)
#import resource
#resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])
#import threading
#threading.stack_size(2**27)
#import sys
#sys.setrecursionlimit(10**7)
#fact=[1]
#for i in range(1,100001):
#    fact.append((fact[-1]*i)%mod)
#ifact=[0]*100001
#ifact[100000]=pow(fact[100000],mod-2,mod)
#for i in range(100000,0,-1):
#    ifact[i-1]=(i*ifact[i])%mod
from sys import stdin, stdout
import bisect
from bisect import bisect_left as bl              #c++ lowerbound bl(array,element)
from bisect import bisect_right as br             #c++ upperbound
import itertools
import collections
import math
import heapq
#from random import randint as rn
def modinv(n,p):
    return pow(n,p-2,p)
def ncr(n,r,p):                        #for using this uncomment the lines calculating fact and ifact
    t=((fact[n])*((ifact[r]*ifact[n-r])%p))%p
    return t
def GCD(x,y):
    while(y):
        x, y = y, x % y
    return x
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input().strip()
"""*******************************************************"""
def main():
    k=int(sin())
    g=k*2
    t=0
    h=g
    while(h>0):
        h/=2
        t+=1
    q=2**t-1
    z=q^k
    print 3,3
    print q,z,q
    print k,q,q
    print q,q,k
######## Python 2 and 3 footer by Pajenegod and c1729
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
       ...