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 java.util.Arrays;

public class D_Rnd722 
{
	static long MOD = 998244353;
	static long BLANK = -1_000_000_000;
	
	static int[] lp;
	
	public static void main(String[] args) throws NumberFormatException, IOException
	{
		BufferedReader scan = new BufferedReader(new InputStreamReader(System.in));

		int n = Integer.parseInt(scan.readLine()) * 2;
		lp = primeSieve(n);
		
		long ans = solve(n);
		System.out.println(ans);
	}

	private static long solve(int n) 
	{
		long[] dp = new long[n + 1];
		Arrays.fill(dp, BLANK);
		
		// fill default values
		dp[0] = 1;
		
		for(int i = 1; i <= n; i += 2)
			dp[i] = 0;
		
		return dp(n, dp) % MOD;
	}

	/*private static long dp(int size, long[] dp) 
	{
		if(dp[size] != BLANK) return dp[size];
		
		dp[size] = 0;
		int left = size - 2;
		
		// get all the possible sizes with stuff in between
		while(left > 0)
		{
			dp[size] += dp(left, dp);
			left -= 2;
		}
		
		int ringSize = 2;
		while(ringSize + ringSize - 2 <= size)
		{
			if(size % (ringSize + ringSize - 2) == 0)
				dp[size]++;
			
			ringSize ++;
		}
		
		dp[size] %= MOD;
		return dp[size];
	}*/
	
	public static long dp(int n, long[] dp)
	{
		long totalSoFar = 0;
		
		for(int size = 2; size <= n; size += 2)
		{
			dp[size] = totalSoFar;
			
			dp[size] += getNumberUniqueFactors(size);
			
			dp[size] %= MOD;
			totalSoFar =  (totalSoFar + dp[size]) % MOD;
		}
		
		return dp[n];
	}

	private static long getNumberUniqueFactors(int size) 
	{
		/*size /= 2;
		long count = 1;
		if(size != 1) count++;
		
		for(int i = 2; i * i <= size; i++)
		{
			if(size % i == 0)
			{
				count ++;
				if(i * i != size) count++;
			}
		}
		
		return count;
		*/
		
		size /= 2;
		
		long ans = 1;
		long term = 0;
		int factor;
		
		while(size > 1)
		{
			term = 0;
			factor = lp[size];
			
			while(size % factor == 0)
			{
				term ++;
				size /= factor;
			}
			
			ans *= (term + 1);
		}
		
		return ans;
		
	}
	
	// Linear prime sieve, but forces you to save the smallest prime factors (more memory)
	// Useful when you need to factor numbers in a large range, but not all of them (like the above function)
	static int[] primeSieve(int N) {
	  int[] lp = new int[N + 1];
	  int[] pr = new int[N + 1]; // could shrink this to the number of primes
	  int prSize = 0;
	  for (int i = 2; i <= N; ++i) {
	    if (lp[i] == 0) {
	      lp[i] = i;
	      pr[prSize++] = i;
	    }
	    int ub = Math.min(lp[i], N / i);
	    for (int j = 0; j < prSize && pr[j] <= ub; ++j) {
	      lp[i * pr[j]] = pr[j];
	    }
	  }
	  return lp;
	}
	
	
}