Codeforces Round #645 (Div. 2) D. The Best Vacation

题目链接
You’ve been in love with Coronavirus-chan for a long time, but you didn’t know where she lived until now. And just now you found out that she lives in a faraway place called Naha.

You immediately decided to take a vacation and visit Coronavirus-chan. Your vacation lasts exactly x days and that’s the exact number of days you will spend visiting your friend. You will spend exactly x consecutive (successive) days visiting Coronavirus-chan.

They use a very unusual calendar in Naha: there are n months in a year, i-th month lasts exactly di days. Days in the i-th month are numbered from 1 to di. There are no leap years in Naha.

The mood of Coronavirus-chan (and, accordingly, her desire to hug you) depends on the number of the day in a month. In particular, you get j hugs if you visit Coronavirus-chan on the j-th day of the month.

You know about this feature of your friend and want to plan your trip to get as many hugs as possible (and then maybe you can win the heart of Coronavirus-chan).

Please note that your trip should not necessarily begin and end in the same year.

Input

The first line of input contains two integers n and x (1≤n≤2e5) — the number of months in the year and the number of days you can spend with your friend.

The second line contains n integers d1,d2,…,dn, di is the number of days in the i-th month (1≤di≤1e6).

It is guaranteed that 1≤x≤d1+d2+…+dn.

Output

Print one integer — the maximum number of hugs that you can get from Coronavirus-chan during the best vacation in your life.

Examples

input

3 2
1 3 1

output

input

3 6
3 3 3

output

input

5 6
4 2 3 1 3

output

典型的前缀和加二分~
题意就是在这些天里面找一段连续的天数 xxx，使得总和最大。因为每年的天数是循环的，所以答案肯定就出现在两年内~
用 pre1pre1pre1 记录数组 ddd 的前缀和，对每一个 ddd，总拥抱数即为 d∗(d+1)/2d*(d+1)/2d∗(d+1)/2，我们再用 pre2pre2pre2 记录持续天数的前缀和，然后遍历计算即可~
对每一个大于 xxx 的前缀和 pre1[i+1]pre1[i+1]pre1[i+1]，我们二分查找出大于 pre−xpre-xpre−x 的前缀和的位置 pospospos，那么现有的总拥抱数即为 pre2[i+1]−pre2[pos]pre2[i+1]-pre2[pos]pre2[i+1]−pre2[pos]，总天数 daydayday 即为pre1[i+1]−pre1[pos]pre1[i+1]-pre1[pos]pre1[i+1]−pre1[pos]，但此时 ddd 可能小于 xxx，所以我们要补上少的这部分，通过计算不难发现这部分即为 d[pos−1]∗(d[pos−1]+1)/2−(d[pos−1]−res)∗(d[pos−1]−res+1)/2d[pos-1]*(d[pos-1]+1)/2-(d[pos-1]-res)*(d[pos-1]-res+1)/2d[pos−1]∗(d[pos−1]+1)/2−(d[pos−1]−res)∗(d[pos−1]−res+1)/2，每次更新一下答案即可，AC代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
main(){ll n,x,ans=0;cin>>n>>x;vector<ll>d(2*n),pre1={0},pre2={0};for(ll i=0;i<n;i++) cin>>d[i],d[n+i]=d[i];for(ll i=0;i<2*n;i++) pre1.push_back(pre1.back()+d[i]),pre2.push_back(pre2.back()+d[i]*(d[i]+1)/2);for(ll i=0;i<2*n;i++){if(pre1[i+1]>=x){ll pos=upper_bound(pre1.begin(),pre1.end(),pre1[i+1]-x)-pre1.begin();ll cnt=pre2[i+1]-pre2[pos];ll day=pre1[i+1]-pre1[pos];ll res=x-day;cnt+=d[pos-1]*(d[pos-1]+1)/2-(d[pos-1]-res)*(d[pos-1]-res+1)/2;ans=max(ans,cnt);}}cout<<ans;
}