28237
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 81367
 8554
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 39004
 22308
 16054
 71468
 60501
 100
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 59632
 90783
 5611
 1000
 69475
 69753
 10
 5
 582
 8
 36961
 5071
 20
 53213
 92775
 28738
 100000
 92280
 37944
 9
 2627
 7
 3

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    #include <bits/stdc++.h>
    #include <ext/pb_ds/assoc_container.hpp>
    #include <ext/pb_ds/tree_policy.hpp>
    #include <ext/pb_ds/detail/standard_policies.hpp>
 
    using namespace std;
    using namespace __gnu_pbds;
 
    #define endl "\n" //interactive 
    #define fi first
    #define se second
    #define pb push_back
    #define int long long   
    #define ll long long
    #define ld long double
    #define pii pair<ll,ll>
    typedef vector<int> vi;
    typedef vector<pair<int,int>> vpi;
    typedef vector<vector<int>> vvi;
 
    #define read(t) ll t;cin>>t;
    #define all(value) value.begin(),value.end()
    #define f(a,b,c) for(ll a=b;a<c;a++)
    #define For(i,n) for(ll i=0;i<n;i++)
 
    #define deb(value) cerr << #value << " is =" << value << endl;
    #define pr(value) cerr<<value<<" ";
    #define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL) ;
 
    typedef tree<pair<ll,ll>, null_type,less<pair<ll,ll>>, rb_tree_tag,tree_order_statistics_node_update> omultiset; // store ind on sec 
    typedef tree<long long,null_type,less<long long>,rb_tree_tag,tree_order_statistics_node_update> oset;
    typedef tree<long long,null_type,greater<long long>,rb_tree_tag,tree_order_statistics_node_update> oset1; // decreasing 
   /* member functions :
                1. order_of_key(k) : number of elements strictly lesser than k
                2. find_by_order(k) : k-th element in the set (counting from zero)*/
 
    //a.erase(unique(a.begin(),a.end()),a.end()); 
 
    double pi=3.141592653589793238462641;
    ll dx[4]={1,-1,0,0},dy[4]={0,0,1,-1};
 
    const int  mod  =  998244353 ; 
    const int  mod2 =  998244353;
    const int  inf  =  1e18;
 
    long long binpow(int a,int b,int m)
    {   
        a %= m;int res = 1;
        while (b > 0)
        { 
            if (b & 1)res=(res*a)%m;
            a=(a*a)%m  ;    b/=2; 
        }
        return res;
    }
    int modinv(int n,int m) {return binpow(n,m-2,m);}
 
/*
    # don't apply unnecessary operations 
    # do check boundary cases
    # change approach / find another way  
    # Don't stuck! :- * re-solve 
                      * check for another sol
                      * check function , cal 
*/
/*......................................................................................................................*/
 
const int N=1e5+5;
int prime[N];

void solve() 
{   
    int n;
    cin>>n;
    for(int i=2;i<=n;i++)
    {
        
        if(!prime[i])
        {
            prime[i]=0;
            for(int j=i+i;j<=n;j+=i)prime[j]+=j/i;
        }
    }
    int ans=0;
    for(int i=2;i<=n;i++)
    {
        ans+=prime[i];
    }
    cout<<ans*4<<endl;
}
 
signed main() 
{
    fast;
    int test=1;
    // cin>>test;
    while(test--)solve();
 
    return 0;
}   