Problem: Kavi has 2n points lying on the OX axis, i-th of which is located at x = i.

Kavi considers all ways to split these 2n points into n pairs. Among those, he is interested in good pairings, which are defined as follows:

Consider n segments with ends at the points in correspondent pairs. The pairing is called good, if for every 2 different segments A and B among those, at least one of the following holds:

  * One of the segments A and B lies completely inside the other. 
  * A and B have the same length. 



Consider the following example:

<image>

A is a good pairing since the red segment lies completely inside the blue segment.

B is a good pairing since the red and the blue segment have the same length.

C is not a good pairing since none of the red or blue segments lies inside the other, neither do they have the same size.

Kavi is interested in the number of good pairings, so he wants you to find it for him. As the result can be large, find this number modulo 998244353.

Two pairings are called different, if some two points are in one pair in some pairing and in different pairs in another.

Input

The single line of the input contains a single integer n (1≤ n ≤ 10^6).

Output

Print the number of good pairings modulo 998244353.

Examples

Input


1


Output


1

Input


2


Output


3

Input


3


Output


6

Input


100


Output


688750769

Note

The good pairings for the second example are: 

<image>

In the third example, the good pairings are: 

<image>

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		
		public Pair(int first, int second) 
		{
			this.first = first;
			this.second = second;
		}
		
		public int compareTo(Pair ob) 
		{
			return (int)(ob.second - second);
		}
	}
	
	
	static class Tuple implements Comparable<Tuple>
	{
		int first, second,third;
		public Tuple(int first, int second, int third)
		{
			this.first = first;
			this.second = second;
			this.third = third;
		}
		public int compareTo(Tuple o)
		{
			return (int)(o.third - this.third);
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static long lcm(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		//Count the number of coprime's upto N
		public static long phi(long n)  //euler totient/phi function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }	
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) 
	    {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(long n, long k) 
	    {
			long ans = 1;
			for(long i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   public static boolean isPrime(long n) 
			{
				if(n == 1)
				{
					return false;
				}
				for(int i = 2;i*i<=n;i++)
				{
					if(n%i == 0)
					{
						return false;
					}
				}
				return true;
			}
			
			public static List<Long> SieveList(int n)
			{
				boolean prime[] = new boolean[(int)(n+1)];
				Arrays.fill(prime, true);
				
				List<Long> l = new ArrayList<>();
				for (long p=2; p*p<=n; p++) 
				{ 
					if (prime[(int)(p)] == true) 
					{ 	
					    for(long i=p*p; i<=n; i += p) 
						{
						    prime[(int)(p)] = false; 
						}	
					} 
				} 
		 
				for (long p=2; p<=n; p++) 
				{
				    if (prime[(int)(p)] == true)
				    {
				       l.add(p); 
				    }
				}
				
				return l;
			}

			public static long log2(long n)
		    {
				 long ans = (long)(log(n)/log(2));
				 return ans;
			}
			   
			public static boolean isPow2(long n)
			{
				 return (n  != 0 && ((n & (n-1))) == 0);
			}	
		   
		   public static boolean isSq(int x)  
		   {
			    long s = (long)Math.round(Math.sqrt(x));
			    return s*s==x;
		   }
		
			
			   /*
			    * 
			    * >=         <=
			   0 1 2 3 4 5 6 7  
			   5 5 5 6 6 6 7 7
			   lower_bound for 6 at index 3 (>=)
			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
			   */

		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1, r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
			
			public static void Sort(long[] a) 
			{
				List<Long> l = new ArrayList<>();
				for (long i : a) l.add(i);
				Collections.sort(l);
//				Collections.reverse(l);  //Use to Sort decreasingly 
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
			
			public static void ssort(char[] a)
			{
				List<Character> l = new ArrayList<>();
				for (char i : a) l.add(i);
				Collections.sort(l);
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
		
	
	public static void main(String[] args) throws Exception 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int[] dp = new int[1000005];
		
		for(int i = 1;i <= 1000000;i++)
		{
			for(int j = i;j <= 1000000;j+=i)
			{
				dp[j]++;
			}
		}
		
		long ans = 0;
		for(int i = 1;i<n;i++)
		{
			ans = ans * 2 + dp[i];
			ans %= temp;
		}
		
		ans += dp[n];
		ans %= temp;
		fout.println(ans);
		    
		fout.close();
	}
}