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A. Heating

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputSeveral days ago you bought a new house and now you are planning to start a renovation. Since winters in your region can be very cold you need to decide how to heat rooms in your house.

Your house has $$$n$$$ rooms. In the $$$i$$$-th room you can install at most $$$c_i$$$ heating radiators. Each radiator can have several sections, but the cost of the radiator with $$$k$$$ sections is equal to $$$k^2$$$ burles.

Since rooms can have different sizes, you calculated that you need at least $$$sum_i$$$ sections in total in the $$$i$$$-th room.

For each room calculate the minimum cost to install at most $$$c_i$$$ radiators with total number of sections not less than $$$sum_i$$$.

Input

The first line contains single integer $$$n$$$ ($$$1 \le n \le 1000$$$) — the number of rooms.

Each of the next $$$n$$$ lines contains the description of some room. The $$$i$$$-th line contains two integers $$$c_i$$$ and $$$sum_i$$$ ($$$1 \le c_i, sum_i \le 10^4$$$) — the maximum number of radiators and the minimum total number of sections in the $$$i$$$-th room, respectively.

Output

For each room print one integer — the minimum possible cost to install at most $$$c_i$$$ radiators with total number of sections not less than $$$sum_i$$$.

Example

Input

4 1 10000 10000 1 2 6 4 6

Output

100000000 1 18 10

Note

In the first room, you can install only one radiator, so it's optimal to use the radiator with $$$sum_1$$$ sections. The cost of the radiator is equal to $$$(10^4)^2 = 10^8$$$.

In the second room, you can install up to $$$10^4$$$ radiators, but since you need only one section in total, it's optimal to buy one radiator with one section.

In the third room, there $$$7$$$ variants to install radiators: $$$[6, 0]$$$, $$$[5, 1]$$$, $$$[4, 2]$$$, $$$[3, 3]$$$, $$$[2, 4]$$$, $$$[1, 5]$$$, $$$[0, 6]$$$. The optimal variant is $$$[3, 3]$$$ and it costs $$$3^2+ 3^2 = 18$$$.

Codeforces (c) Copyright 2010-2020 Mike Mirzayanov

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