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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
#define min(a,b) ((a)>(b)?(b):(a))
const int maxn=1e6+10;
int n;
const int P=998244353;
ll dp[maxn];
int is[maxn],fac[maxn],prime[maxn],cnt,pre[maxn];
void factor() {
    int n=maxn-1;
	is[1]=1;fac[1]=1;
	for(int i=1;i<=n;i++){
		if(!is[i]){prime[++cnt]=i;fac[i]=2;pre[i]=1;}
		for(int j=1;j<=cnt&&i*prime[j]<=n;j++) {
			is[prime[j]*i]=1;
			if(i%prime[j]){
				fac[i*prime[j]]=2*fac[i];
				pre[i*prime[j]]=1;
			} else {
				pre[i*prime[j]]=pre[i]+1;
				fac[i*prime[j]]=fac[i]/(pre[i]+1)*(pre[i]+2);
				break;
			}
		}
	}
}
int main() {
    factor();
    scanf("%d",&n);
    dp[0]=dp[1]=1;
    ll sum=2;
    for(int i=2;i<=n;++i) {
        dp[i]=(dp[i]+sum)%P;
        dp[i]=(dp[i]+(fac[i]-1))%P;
        sum=(dp[i]+sum)%P;
    }
    printf("%lld\n",dp[n]);
}