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 Fortran (gfortran v4.8.4) (judge verdict: Runtime Error)

program test
implicit none
integer(8) :: i=0,j=0,n=0,k=0,t(100000)=0,d(100000)=0,dummy(100000)=0,num=0
integer(8) :: tt(100000)=0,dd(100000)=0,lost(100000)=0,lost_cum(100000)=0
integer(8) :: get(100000)=0,get_cum(100000)=0,total(100000)=0,sumi_a(100000)=0,sumi_b(100000)=0
integer(8) :: kind_max_a(100000)=0,kind_max_b(100000)=0,kind_count=0,lost_count=0,get_count=0,count=0,error

read(*,*) n,k
do i=1,n
	read(*,*) t(i),d(i)
enddo

if(n==k) then
	goto 100
else
	!まずは降順に並べる
	call heapsort_down(n,d(1:n),t(1:n))
	
	!1～k番目が暫定食べる、k+1～n番目が食べない
	tt(1:n) = t(1:n)
	dd(1:n) = d(1:n)
	
	!食べるグループで、各種類の最高値を計算
	do i=1,k
		kind_max_a(tt(i)) = max(kind_max_a(tt(i)),dd(i))
	enddo
	
	!各種類の最高値を1個だけ残して、それ以外は新たな種類の寿司に置換される可能性がある
	lost_count = 0
	do i=1,k
		if(dd(i) == kind_max_a(tt(i)) .and. sumi_a(tt(i)) == 0 ) then
			sumi_a(tt(i)) = 1
		else
			lost_count=lost_count+1
			lost(lost_count) = dd(i)
		endif
	enddo
	
	if(lost_count == 0)then
		goto 200
	endif
	
	!昇順ソートをかける
	call heapsort_up(lost_count,lost(1:lost_count),dummy(1:lost_count))
	
	!失う価値の累積和
	do i=1,lost_count
		if(i == 1) then
			lost_cum(i) = lost(1)
		else
			lost_cum(i) = lost_cum(i-1) + lost(i)
		endif
	enddo
	
	!食べるグループの種類は何種類？
	call heapsort_up(k,tt(1:k),dd(1:k))
	kind_count = 1
	if(k>=2)then
		do i =2,k
			if(tt(i) .ne. tt(i-1))then
			 kind_count = kind_count + 1
			endif
		enddo
	endif
	
	!食べないグループで、各種類の最高値を列挙
	do i=k+1,n
		kind_max_b(tt(i)) = max(kind_max_b(tt(i)),dd(i))
	enddo
	
	!食べないグループで、各種類の最高値にあたり、かつ食べるグループにない寿司が置換に使われる可能性がある
	get_count = 0
	do i=k+1,n
		if((dd(i) == kind_max_b(tt(i))) .and. kind_max_a(tt(i)) == 0 .and. sumi_b(tt(i)) == 0) then
			get_count = get_count + 1
			get(get_count) = dd(i)
			sumi_b(tt(i)) = 1
		endif
	enddo
	
	if(get_count == 0)then
		goto 300
	endif
	
	!降順ソートをかける
	call heapsort_down(get_count,get(1:get_count),dummy(1:get_count))
	
	!得る価値
	do i=1,get_count
		if(i == 1) then
			get_cum(i) = get(1) + 2*(kind_count+i)-1
		else
			get_cum(i) = get_cum(i-1) + get(i) +  2*(kind_count+i)-1
		endif
	enddo
	
	count = min(lost_count,get_count)
	
	do i = 1,count
		total(i) = get_cum(i) - lost_cum(i)
	enddo
	
	write(*,*) sum(d(1:k))+ kind_count*kind_count+ max(0,maxval(total(1:count)))
endif

if(1 == 0)then
100 continue
	call heapsort_up(n,t(1:k),d(1:k))
	num = 1
	if(k>=2)then
		do i =2,k
			if(t(i) .ne. t(i-1))then
			 num = num + 1
			endif
		enddo
	endif
	write(*,*) sum(d(1:k)) + num*num
endif

if(1 == 0)then
200 continue
	call heapsort_up(n,t(1:k),d(1:k))
	num = 1
	if(k>=2)then
		do i =2,k
			if(t(i) .ne. t(i-1))then
			 num = num + 1
			endif
		enddo
	endif
	write(*,*) sum(d(1:k)) + num*num
endif

if(1 == 0)then
300 continue
	call heapsort_up(n,t(1:k),d(1:k))
	num = 1
	if(k>=2)then
		do i =2,k
			if(t(i) .ne. t(i-1))then
			 num = num + 1
			endif
		enddo
	endif
	write(*,*) sum(d(1:k)) + num*num + d(1000000) !ERROR
endif

contains
 
subroutine heapsort_down(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) > y1(j+1)) j = j+1
				endif
				
				if (t1 > y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_down

subroutine heapsort_up(n,y1,y2)
 
	implicit none
	integer(8),intent(in) :: n
	integer(8),intent(inout) :: y1(1:n)
	integer(8),intent(inout) :: y2(1:n)
	integer(8) :: i,k,j,l
	integer(8) :: t1,t2
	
	l = n/2+1
	k = n
 
	do while (k /= 1)
		 if(l > 1) then
				l = l-1
				t1 = y1(l)
				t2 = y2(l)
				
		 else
				t1 = y1(k)
				t2 = y2(k)
				y1(k) = y1(1)
				y2(k) = y2(1)
				k = k-1
				if(k == 1) then
					 y1(1) = t1
					 y2(1) = t2
					 exit
				endif
		 endif
 
		 i = l
		 j = l+l
 
		 do while(j <= k)
				if(j < k) then
					 if(y1(j) < y1(j+1)) j = j+1
				endif
				
				if (t1 < y1(j)) then
					 y1(i) = y1(j)
					 y2(i) = y2(j)
					 i = j
					 j = j+j
				else
					 j = k+1
				endif
		 
		 enddo
		 
		 y1(i) = t1
		 y2(i) = t2
	
	enddo
	return
 
end subroutine heapsort_up

end program