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.
from __future__ import division, print_function
# import threading
# threading.stack_size(2**27)
# import sys
# sys.setrecursionlimit(10**6)
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input())
import math  
def Divisors(n) : 
    l = []  
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                l.append(n//i)
    return l
q=[]
def dfs(n,d,v,c):
    global q
    v[n]=1
    x=d[n]
    q.append(n)
    j=c
    for i in x:
        if i not in v:
            f=dfs(i,d,v,c+1)
            j=max(j,f)
            # print(f)
    return j
def primeFactors(n): 
    a=set()
    while n % 2 == 0:
        a.add(2)  
        n = n // 2
    for i in range(3,int(math.sqrt(n))+1,2): 
        while n % i== 0: 
            a.add(i) 
            n = n // i 
    if n > 2: 
        a.add(n)
    return a

"""*******************************************************"""
def main():
    n=inin()
    print("2 3")
    a=[[0,0,0],[0,0,0]]
    a[0][1]=n
    x=0
    y=n
    while(n>0):
        n=n//2
        x+=1
    z=pow(2,x)
    a[1][0]=z
    a[1][1]=z+y
    a[0][0]=z+y
    a[1][2]=y
    for i in a:
        print(*i)
    

######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
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])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
 
# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

# threading.Thread(target=main).start()
if __name__== "__main__":
  main()