Problem:
Score : 400 points

Problem Statement

There are N pieces of sushi. Each piece has two parameters: "kind of topping" t_i and "deliciousness" d_i.
You are choosing K among these N pieces to eat.
Your "satisfaction" here will be calculated as follows:

The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".

The base total deliciousness is the sum of the deliciousness of the pieces you eat.

The variety bonus is x*x, where x is the number of different kinds of toppings of the pieces you eat.

You want to have as much satisfaction as possible.
Find this maximum satisfaction.

Constraints

1 \leq K \leq N \leq 10^5

1 \leq t_i \leq N

1 \leq d_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
t_1 d_1
t_2 d_2
.
.
.
t_N d_N

Output

Print the maximum satisfaction that you can obtain.

Sample Input 1

5 3
1 9
1 7
2 6
2 5
3 1

Sample Output 1

26

If you eat Sushi 1,2 and 3:

The base total deliciousness is 9+7+6=22.

The variety bonus is 2*2=4.

Thus, your satisfaction will be 26, which is optimal.

Sample Input 2

7 4
1 1
2 1
3 1
4 6
4 5
4 5
4 5

Sample Output 2

25

It is optimal to eat Sushi 1,2,3 and 4.

Sample Input 3

6 5
5 1000000000
2 990000000
3 980000000
6 970000000
6 960000000
4 950000000

Sample Output 3

4900000016

Note that the output may not fit into a 32-bit integer type.

Solution in Lua (5.3.2) (judge verdict: Runtime Error)

local N, K = io.read("n", "n")
local sushi_list = {}
--local sushi_types = {}
for i=1,N do
    local t, d = io.read("n", "n")
    local tuple = {t, d, false} -- 3rd element: used flag
    --local tuple = {t, d}
    sushi_list[i] = tuple
    --if not sushi_types[t] then
    --    sushi_types[t] = {}
    --end
    --table.insert(sushi_types[t], tuple)
end

-- sort order by deliciousness (most tasty sushi first)
table.sort(sushi_list, function (x, y) return x[2] > y[2] end)
--for _,sushi_t in pairs(sushi_types) do
--    table.sort(sushi_t, function (x, y) return x[2] > y[2] end)
--end

-- maximaize base score
local base_score = 0
local used = {}
for i=1,K do
    local tuple = sushi_list[i]
    base_score = base_score + tuple[2]
    if used[tuple[1]] == nil then
        used[tuple[1]] = 0
    end
    used[tuple[1]] = used[tuple[1]] + 1
    tuple[3] = true
end
local not_used_sushis = {}
local not_used_sushis_appeared = {}
for i=K+1,N do
    local tuple = sushi_list[i]
    if used[tuple[1]] == nil then
        if not_used_sushis_appeared[tuple[1]] then
            -- never use this one
        else
            not_used_sushis_appeared[tuple[1]] = true
            table.insert(not_used_sushis, tuple)
        end
    end
end
local ntypes = 0
for k,_ in pairs(used) do
    ntypes = ntypes + 1
end
local bonus_score = ntypes * ntypes
local score = base_score + bonus_score


if false then
    print("maximized base")
    for i,v in ipairs(sushi_list) do
        print(v[1],v[2],v[3])
    end
    print("not used sushis")
    for i,v in ipairs(not_used_sushis) do
        print(v[1],v[2],v[3])
    end
    print("--")
end


-- try increase bonus
local discard_i = K
local take_i = 1
local max_score = score
while true do
    -- discard least tasty one (constraint: keep ntypes)
    local discard_tuple = sushi_list[discard_i]
    if used[discard_tuple[1]] >= 2 then
        used[discard_tuple[1]] = used[discard_tuple[1]] - 1
        local take_tuple = not_used_sushis[take_i]
        take_i = take_i + 1
        local score_temp = (score - discard_tuple[2] + take_tuple[2]) - ntypes * ntypes + (ntypes + 1) * (ntypes + 1)
        if score_temp > max_score then
            max_score = score_temp
        end
        discard_tuple[3] = false
        take_tuple[3] = true
        score = score_temp
        ntypes = ntypes + 1
        if take_i > #not_used_sushis then
            break
        end
    else
        -- keep ntypes
    end
    discard_i = discard_i - 1
    if discard_i == 0 then
        break
    end
end

print(max_score)