Apart from the number and

Title Description

Given a positive integer n and k, calc j (n, k) = k mod 1 + k mod 2 + k mod 3 + ... + k mod n in
where MOD k i represents the remainder of i divided by k.
For example j (5, 3) = 3 mod 1 + 3 mod 2 + 3 mod 3 + 3 mod 4 + 3 mod 5 = 0 + 1 + 0 + 3 + 3 = 7

Entry

Only the input line, comprising two integers n, k.
1 <= n, k <= 10 ^ 9

Export

Output only one row, i.e., j (n, k).

Sample input

5 3

Sample Output

7

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
ll n,k,ans;
int main()
{
    scanf("%lld%lld",&n,&k);
    ans=n*k;
    for (ll l=1,r=0; l<=n; l=r+1)
    {
        if (k/l)
        {
            r=min(n,k/(k/l));
        }
        else
        {
            r=n;
        }
        ans-=(k/l)*(r-l+1)*(l+r)>>1;
    }
    printf("%lld\n",ans);
}