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 LuaJIT (2.0.4) (judge verdict: Accepted)

local mod = 998244353
local mma, mmi = math.max, math.min
local mfl = math.floor

local function bmul(x, y)
  local x0, y0 = x % 31596, y % 31596
  local x1, y1 = mfl(x / 31596), mfl(y / 31596)
  return (x1 * y1 * 62863 + (x1 * y0 + x0 * y1) * 31596 + x0 * y0) % mod
end

local function badd(x, y)
  return (x + y) % mod
end

local function bsub(x, y)
  return x < y and x - y + mod or x - y
end

local function modpow(src, pow)
  local res = 1
  while 0 < pow do
    if pow % 2 == 1 then
      res = bmul(res, src)
      pow = pow - 1
    end
    src = bmul(src, src)
    pow = mfl(pow / 2)
  end
  return res
end

local n = io.read("*n")
local a = {}
local tot = 0
for i = 1, n do
  a[i] = io.read("*n")
  tot = tot + a[i]
end
local htot = mfl(tot / 2)
local dp = {}
local single = {}
local curmax = 0
for i = 1, htot do
  dp[i] = 0
  single[i] = 0
end
for i = 1, n do
  local ai = a[i]
  local lim = mmi(curmax, htot - ai)
  for j = lim, 1, -1 do
    dp[j + ai] = badd(dp[j + ai], bmul(2, dp[j]))
    single[j + ai] = badd(single[j + ai], single[j])
  end
  if ai <= htot then
    dp[ai] = badd(dp[ai], 2)
    single[ai] = badd(single[ai], 1)
  end
  curmax = curmax + ai
end
local ret = modpow(3, n)
if tot % 2 == 0 then
  local rmsum = 1
  for i = 1, htot do
    rmsum = badd(rmsum, dp[i])
  end
  rmsum = bsub(rmsum, single[htot])
  rmsum = bmul(rmsum, 3)
  ret = bsub(ret, rmsum)
else
  local rmsum = 1
  for i = 1, htot do
    rmsum = badd(rmsum, dp[i])
  end
  rmsum = bmul(rmsum, 3)
  ret = bsub(ret, rmsum)
end
print(ret)