题意：给出n和k，1≤n,k≤1e9，计算
切入点是k/i 和 k/(i+1)差距不大。令pi = k/i, ri = k%i。如果pi+1 == pi，那么ri+1 == k - pi(i+1) == ri - pi，
对于pi+z == pi，ri+z == ri - z*pi，这是等差数列可以O(1)计算出来。如果上界为j，那么k/j ≤ pi，j ≤ k/pi。
/*********************************************************
* --------------Tyrannosaurus--------- *
* author AbyssalFish *
**********************************************************/
#include<bits/stdc++.h>
using namespace std; typedef long long ll; ll solve(int n, int k)
{
ll ans = ;
for(int i = ; i <= n; ){
int p = k/i;
if(!p){
ans += (n+-i)*(ll)k;
break;
}
int j = min(k/p, n), r = k-i*p;
ans += (2LL*r - (j-i)*p)*(j-i+1LL)/2LL;
i = j+;
}
return ans;
} //#define LOCAL
int main()
{
#ifdef LOCAL
freopen("in.txt","r",stdin);
#endif
int n, k;
while(~scanf("%d%d",&n,&k)){
printf("%lld\n", solve(n,k));
}
return ;
}