Problem: Attention: we lost all the test cases for this problem, so instead of solving the problem, we need you to generate test cases. We're going to give you the answer, and you need to print a test case that produces the given answer. The original problem is in the following paragraph.

People don't use cash as often as they used to. Having a credit card solves some of the hassles of cash, such as having to receive change when you can't form the exact amount of money needed to purchase an item. Typically cashiers will give you as few coins as possible in change, but they don't have to. For example, if your change is 30 cents, a cashier could give you a 5 cent piece and a 25 cent piece, or they could give you three 10 cent pieces, or ten 1 cent pieces, two 5 cent pieces, and one 10 cent piece. Altogether there are 18 different ways to make 30 cents using only 1 cent pieces, 5 cent pieces, 10 cent pieces, and 25 cent pieces. Two ways are considered different if they contain a different number of at least one type of coin. Given the denominations of the coins and an amount of change to be made, how many different ways are there to make change?

As we mentioned before, we lost all the test cases for this problem, so we're actually going to give you the number of ways, and want you to produce a test case for which the number of ways is the given number. There could be many ways to achieve this (we guarantee there's always at least one), so you can print any, as long as it meets the constraints described below.

Input

Input will consist of a single integer A (1 ≤ A ≤ 105), the desired number of ways.

Output

In the first line print integers N and M (1 ≤ N ≤ 106, 1 ≤ M ≤ 10), the amount of change to be made, and the number of denominations, respectively.

Then print M integers D1, D2, ..., DM (1 ≤ Di ≤ 106), the denominations of the coins. All denominations must be distinct: for any i ≠ j we must have Di ≠ Dj.

If there are multiple tests, print any of them. You can print denominations in atbitrary order.

Examples

Input

18


Output

30 4
1 5 10 25


Input

3


Output

20 2
5 2


Input

314


Output

183 4
6 5 2 139

Solution (in JAVA):

Solution:
/**
 * DA-IICT
 * Author : PARTH PATEL
 */
import java.io.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;


public class B867 
{

	public static int mod = 1000000007;
	static FasterScanner in = new FasterScanner();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) 
	{
		int n=in.nextInt();
		out.println((2*n-1)+" "+2);
		out.println("1 2");
		out.close();

	}
	

	/////////////////SEGMENT TREE (BUILD-UPDATE-QUERY)/////////////////////////////
	/////////////////UPDATE FOLLOWING METHODS AS PER NEED//////////////////////////
	/*
	public static void buildsegmenttree(int node,int start,int end)
	{
		if(start==end)
		{
			// Leaf node will have a single element
			segmenttree[node]=arr[start];
		}
		else
		{
			int mid=start+(end-start)/2;
			// Recurse on the left child
			buildsegmenttree(2*node, start, mid);
			// Recurse on the right child
			buildsegmenttree(2*node+1, mid+1, end);
			// Internal node will have the sum of both of its children
			segmenttree[node]=segmenttree[2*node]+segmenttree[2*node+1];
		}
	}
	
	public static void updatesegmenttree(int node,int start,int end,int idx,int val)
	{
		if(start==end)
		{
			//Leaf Node
			arr[idx]+=val;
			segmenttree[node]+=val;
		}
		else
		{
			int mid=start+(end-start)/2;
			if(start<=idx && idx<=mid)
			{
				// If idx is in the left child, recurse on the left child
				updatesegmenttree(2*node, start, mid, idx, val);
			}
			else
			{
				// if idx is in the right child, recurse on the right child
				updatesegmenttree(2*node+1, mid+1, end, idx, val);
			}
			// Internal node will have the sum of both of its children
			segmenttree[node]=segmenttree[2*node]+segmenttree[2*node+1];
		}
	}
	
	public static long querysegmenttree(int node,int start,int end,int l,int r)
	{
		if(r<start || end<l)
		{
			// range represented by a node is completely outside the given range
			return 0;
		}
		if(l <= start && end <= r)
	    {
	        // range represented by a node is completely inside the given range
	        return segmenttree[node];
	    }
		// range represented by a node is partially inside and partially outside the given range
		int mid=start+(end-start)/2;
		long leftchild=querysegmenttree(2*node, start, mid, l, r);
		long rightchild=querysegmenttree(2*node+1, mid+1, end, l, r);
		return (leftchild+rightchild);
	}
	*/
	
	public static long pow(long x, long n, long mod) 
	{
		long res = 1;
		for (long p = x; n > 0; n >>= 1, p = (p * p) % mod) 
		{
			if ((n & 1) != 0) 
			{
				res = (res * p % mod);
			}
		}
		return res;
	}

	public static long gcd(long n1, long n2) 
	{
		long r;
		while (n2 != 0) 
		{
			r = n1 % n2;
			n1 = n2;
			n2 = r;
		}
		return n1;
	}

	public static long lcm(long n1, long n2) 
	{
		long answer = (n1 * n2) / (gcd(n1, n2));
		return answer;
	}

	static class FasterScanner 
	{
		private byte[] buf = new byte[1024];
		private int curChar;
		private int snumChars;

		public int read() 
		{
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) 
			{
				curChar = 0;
				try 
				{
					snumChars = System.in.read(buf);
				} catch (IOException e) 
				{
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String nextString() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public long nextLong() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
			long res = 0;
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} 
			while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
			int res = 0;
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} 
			while (!isSpaceChar(c));
			return res * sgn;
		}

		public int[] nextIntArray(int n) 
		{
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) 
			{
				arr[i] = nextInt();
			}
			return arr;
		}

		public long[] nextLongArray(int n) 
		{
			long[] arr = new long[n];
			for (int i = 0; i < n; i++)
			{
				arr[i] = nextLong();
			}
			return arr;
		}

		private boolean isSpaceChar(int c) 
		{
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) 
		{
			return c == '\n' || c == '\r' || c == -1;
		}
	}

}