#420 B. Okabe and Banana Trees（Div.2）

题目链接：http://codeforces.com/contest/821/problem/B

B. Okabe and Banana Trees

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Okabe needs bananas for one of his experiments for some strange reason. So he decides to go to the forest and cut banana trees.

Consider the point (x, y) in the 2D plane such that x and y are integers and 0 ≤ x, y. There is a tree in such a point, and it has x + ybananas. There are no trees nor bananas in other points. Now, Okabe draws a line with equation . Okabe can select a single rectangle with axis aligned sides with all points on or under the line and cut all the trees in all points that are inside or on the border of this rectangle and take their bananas. Okabe's rectangle can be degenerate; that is, it can be a line segment or even a point.

Help Okabe and find the maximum number of bananas he can get if he chooses the rectangle wisely.

Okabe is sure that the answer does not exceed 1018. You can trust him.

Input

The first line of input contains two space-separated integers m and b (1 ≤ m ≤ 1000, 1 ≤ b ≤ 10000).

Output

Print the maximum number of bananas Okabe can get from the trees he cuts.

Examples

input

1 5

output

30

input

2 3

output

25

Note

The graph above corresponds to sample test 1. The optimal rectangle is shown in red and has 30 bananas.

解析：昨天12点多才给写出来，刚开始思路错了，找最大矩形暴力枚举矩形面积，可能忽略了一些特殊情况，就算不WA也会TLE，根据函数y轴截距为b，x轴截距为b*m，那么咱枚举y轴截距（比枚举x省时间），最大点即为（-1ll*(y-b)*m， y），然后找最大的，其实每个最大矩形都在那条线上

代码：

#include<bits/stdc++.h>#define N 209using namespace std;//y   b//x   b*mint main(){    int b, m;    scanf("%d%d", &m, &b);    long long ans = 0ll;    for(long long y = 0; y <= b; y++)    {        long long x = -1ll*(y-b)*m;        ans = max(ans, x*(1ll+x)/2ll*(1ll+y)+y*(1ll+y)/2ll*(1ll+x));    }    cout << ans << endl;    return 0;}