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:
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast")

#include <bits/stdc++.h> 

using namespace std;
using i128 = __int128_t;
using u128 = __uint128_t;
using u64 = uint64_t;

//define
#define int long long
#define ll int
#define trav(i,v)   for(auto i: v)
#define rep(i,n)    for(int i=0;i<n;i++)
#define repu(i,k,n) for(int i=k;i<=n;i++)
#define repd(i,k,n) for(int i=k;i>=n;i--)
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define all(v)  v.begin(), v.end()
#define itn int 
#define sz(x) (int)x.size()
    
//typedef
typedef unsigned long long ull;
typedef long double ld;
typedef std::pair<int, int> pii;
typedef std::vector<int> vi;
typedef std::vector< pii > vii;
typedef std::vector< std::vector<int> > vvi;
typedef std::vector< std::pair < pii, int > > vpp;
 
// const long long MOD = 1000000007;
const long long MOD = 998244353;
const long double PI = 3.141592653589793238;
const long long pi = 31415926;
const long long inf = 1000000000000000000;
const long long small_inf = INT_MAX;

int modpow(int x, int n, int mod = MOD){ int res=1; while(n>0){ if(n&1) res=res*x%mod; x=x*x%MOD; n>>=1;} return res;}
int power(int x, int n){ int res=1; while(n>0){ if(n&1) res=res*x; x=x*x; n>>=1; } return res;}

vi fac(2000001,0);

void init(){
    repu(i,2,2000000){
        for(int j=2*i;j<=2000000;j+=2*i){
            fac[j]++;
        }
    }
}

void solve()
{
/*Don't hurry, nothing good comes if you rush*/
    int n;
    cin>>n;
    vi dp(n+1,0);
    dp[0] = 1;
    dp[1] = 1;
    int sum = 2;
    repu(i,2,n){
        dp[i] = sum+fac[2*i];
        sum+=dp[i];
        dp[i]%=MOD;
        sum%=MOD;
    }
    cout<<dp[n]<<endl;
}

signed main(){
    // #ifndef ONLINE_JUDGE
    //    freopen("./input.txt", "r", stdin);
    //    freopen("./output.txt", "w", stdout);
    // #endif
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    int t23 = 1, tt23 = 1; 
    // cin>>t23;
    init();
    while(tt23<=t23)
    {
        // cout<<"Case #"<<tt23<<": ";
        solve();
        tt23++;
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}