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.*;
import java.util.*;

public class Main {
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static StringTokenizer st;
	static String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}
	static int readInt() throws IOException {
		return Integer.parseInt(next());
	}
	static long readLong() throws IOException {
		return Long.parseLong(next());
	}
	static double readDouble() throws IOException {
		return Double.parseDouble(next());
	}
	static char readChar() throws IOException {
		return next().charAt(0);
	}
	static String readLine() throws IOException {
		return br.readLine();
	}
	static class Pair implements Comparable<Pair> {
		int f, s;
		Pair(int f, int s) {
			this.f = f; this.s = s;
		}
		public int compareTo(Pair other) {
			if (this.f != other.f) return this.f - other.f;
			return this.s - other.s;
		}
	}
	final static int mod = 998244353;
	static int sieve[];
	static Map<Integer, Integer> map = new HashMap();
	static int countEvenFactors(int x) {
		map.clear();
		while (sieve[x] != 0) {
			map.put(sieve[x], map.getOrDefault(sieve[x], 0) + 1);
			x /= sieve[x];
		}
		if (x != 1) map.put(x, map.getOrDefault(x, 0) + 1);
		int ret = 1;
		for (int k : map.keySet()) {
			if (k == 2) ret *= map.get(k);
			else ret *= map.get(k) + 1;
		}
		return ret;
	}
	public static void main(String[] args) throws IOException {
		int n = readInt();
		sieve = new int[n<<1|1];
		for (int i = 2; i <= n<<1; ++i) {
			if (sieve[i] == 0) {
				for (int j = i<<1; j <= n<<1; j += i) sieve[j] = i;
			}
		}
		long dp[] = new long[n<<1|1];
		dp[2] = 1;
		for (int i = 4; i <= n<<1; i += 2) {
			dp[i] = countEvenFactors(i);
			dp[i] = (dp[i] + (dp[i - 2]<<1))%mod;
		}
		System.out.println((dp[n<<1] - dp[(n - 1)<<1] + mod)%mod);
	}
}