Problem:
Score : 600 points

Problem Statement

You are given N integers. The i-th integer is a_i.
Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:

Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.

Constraints

3 \leq N \leq 300

1 \leq a_i \leq 300(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
:
a_N

Output

Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.

Sample Input 1

4
1
1
1
2

Sample Output 1

18

We can only paint the integers so that the lengths of the sides of the triangle will be 1, 2 and 2, and there are 18 such ways.

Sample Input 2

6
1
3
2
3
5
2

Sample Output 2

150

Sample Input 3

20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4

Sample Output 3

563038556

Solution in Fortran (gfortran v4.8.4) (judge verdict: Wrong Answer)

PROGRAM ATCODER

implicit none
integer(8) :: n,i,j,count=1,m=998244353,count1=0,count2=0,s
integer(8),allocatable :: a(:),dp(:,:)

read*, n
allocate(a(n))
read*, a(1:n)
allocate(dp(0:n,0:sum(a)))

!! 全ての通り
do i = 1, n, 1
  count = count*3
end do

dp = 0
dp(0,0) = 1
!! dp(t,r)=最初のt個の整数を3色で塗り分け、赤い整数の和がrであるような場合の数
do i = 0, n-1, 1
  do j = 0, sum(a), 1
    !! i+1番目を他の2色で塗る場合
    dp(i+1,j) = mod(dp(i+1,j)+2*dp(i,j),m)
    !! 赤で塗る場合
    if ( j+a(i+1) <= sum(a) ) then
      dp(i+1,j+a(i+1)) = mod(dp(i+1,j+a(i+1))+dp(i,j),m)
    end if
  end do
end do

do j = 1, 3
  do i = ceiling(real(sum(a))/2.d0,8), sum(a), 1
    count1 = mod(count1+dp(n,i),m)
  end do
end do

dp = 0
dp(0,0) = 1
!! dp(t,r)=最初のt個の整数を2色で塗り分け、赤い整数の和がrであるような場合の数
do i = 0, n-1, 1
  do j = 0, sum(a), 1
    !! i+1番目を他の色で塗る場合
    dp(i+1,j) = mod(dp(i+1,j)+dp(i,j),m)
    !! 赤で塗る場合
    if ( j+a(i+1) <= sum(a) ) then
      dp(i+1,j+a(i+1)) = mod(dp(i+1,j+a(i+1))+dp(i,j),m)
    end if
  end do
end do

if (mod(sum(a),2) == 0) then
  do i = 1, 3, 1
    count2 = mod(count2+dp(n,sum(a)/2),m)
  end do
else
  count2 = 0
end if

print'(i0)', count-count1+count2

END PROGRAM ATCODER