Codeforces Round #619 (Div. 2) Ayoub's function

Ayoub thinks that he is a very smart person, so he created a function f(s)f(s) , where ss is a binary string (a string which contains only symbols "0" and "1"). The function f(s)f(s) is equal to the number of substrings in the string ss that contains at least one symbol, that is equal to "1".

More formally, f(s)f(s) is equal to the number of pairs of integers (l,r)(l,r) , such that 1≤l≤r≤|s|1≤l≤r≤|s| (where |s||s| is equal to the length of string ss ), such that at least one of the symbols sl,sl+1,…,srsl,sl+1,…,sr is equal to "1".

For example, if s=s= "01010" then f(s)=12f(s)=12 , because there are 1212 such pairs (l,r)(l,r) : (1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5)(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4),(2,5),(3,4),(3,5),(4,4),(4,5) .

Ayoub also thinks that he is smarter than Mahmoud so he gave him two integers nn and mm and asked him this problem. For all binary strings ss of length nn which contains exactly mm symbols equal to "1", find the maximum value of f(s)f(s) .

Mahmoud couldn't solve the problem so he asked you for help. Can you help him?

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1051≤t≤105 )  — the number of test cases. The description of the test cases follows.

The only line for each test case contains two integers nn , mm (1≤n≤1091≤n≤109 , 0≤m≤n0≤m≤n ) — the length of the string and the number of symbols equal to "1" in it.

Output

For every test case print one integer number — the maximum value of f(s)f(s) over all strings ss of length nn , which has exactly mm symbols, equal to "1".

Example

Input

5
3 1
3 2
3 3
4 0
5 2

Output

4
5
6
0
12

思维题，脑子不好使卡了我半天...问的是如何排列01串里m个1的位置是含有1的子串个数最大。可以逆向思维，用所有子串个数减去全为0的字符串的子串个数。而因为有关系：一个长为k的字符串的子串个数有k(k+1)/2个，故只需要排列好全0字符串即可。
分类讨论：当没有0时或者全为0时答案易得；当1的个数大于等于0的个数时，可以用1把每个0分隔开，这样能保证答案最大，（因为两个单独的0比连续的00所含子串个数少）；当1的个数少于0的个数时，1要尽可能把0分开，m个1最多能分成m+1份，而必须要让0尽可能平均开，所以m+1不能整除n-m（0的个数）的话，要把余数拆成1，尽可能分给别的组，减的时候分别减一下即可。记得开long long。

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long t;
    cin>>t;
    while(t--)
    {
        long long n,m;
        scanf("%lld%lld",&n,&m);
        if(m==0)
        {
            cout<<0<<endl;
            continue;
        }
        if(n==m)
        { 
            cout<<n*(n+1)/2<<endl;
            continue;
        }
        long long ans=(n+1)*n/2;
        long long noz=n-m;
        if(noz<=m)
        {
            ans-=noz;
            cout<<ans<<endl;
            continue;
        }
        long long num =noz/(m+1);//每部分0的个数 
        long long res_group=noz%(m+1);
        ans=ans-(m+1-res_group)*num*(num+1)/2-res_group*(num+1)*(num+2)/2;
        printf("%lld
",ans);
    }
    return 0;
}