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 Lua (5.3.2) (judge verdict: Time Limit Exceeded)

local DBG = false
local function dbgpr(...)
    if DBG then
        io.write("[dbg]")
        print(...)
    end
end
--[[
local function xxxmemoize(f)
    local cache = {}
    return function(i, s)
        if s < 0 then
            return 0
        end
        local k = (i * 100000) + s
        if cache[k] then
            return cache[k]
        end
        local r = f(i, s)
        cache[k] = r
        return r
    end
end
]]

local function memoize(f)
    local cache = {}
    -- http://www.lua.org/doc/jucs05.pdf -- 4 Tables
    -- Make keys contiguous integer so that elements always reside array part
    for i=0, 301 * 90001 do
        cache[i] = -1
    end
    -- i=0,s=0
    cache[0] = 1
    -- i=0,s!=0
    for i=1, 90000 do
        cache[i] = 0
    end
    return function(i, s)
        if s < 0 then
            return 0
        end
        local k = (i * 90001) + s
        if cache[k] >= 0 then
            return cache[k]
        end
        local r = f(i, s)
        cache[k] = r
        return r
    end
end

-- returns (b^n) % m
local function modpow(b, n, m)
    if n == 0 then
        return 1
    --elseif n % 2 == 0 then
    --    return modpow(b * b % m, math.floor(n/2), m) % m
    else
        return b * modpow(b, n - 1, m) % m
    end
end

---
local MOD = 998244353
local N = io.read("*n")
local a = {}
local S = 0
for i=1,N do
    a[i] = io.read("*n")
    S = S + a[i]
end

local combh
combh = memoize(function(i, s)
    --[[
    if i == 0 then
        if s == 0 then
            return 1
        else
            return 0
        end
    end
    ]]
    --print(i, s, a[i])
    return (combh(i - 1, s - a[i]) + combh(i - 1, s)) % MOD
end)

-- ∑A = S/2 and ∑B = S/2 and ∑C = 0
local function getY()
    if S % 2 == 1 then
        return 0
    end
    return combh(N, math.floor(S/2))
end

local combgeh
combgeh = memoize(function(i, s)
    --[[
    if i == 0 then
        if s == 0 then
            return 1
        else
            return 0
        end
    end
    ]]
    return (combgeh(i - 1, s - a[i]) + combgeh(i - 1, s) * 2) % MOD
end)

-- ∑A >= S/2
local function getX()
    local sum = 0
    for s = math.ceil(S/2), S, 1 do
        --local c = combgeh(N, s)
        --dbgpr("s,c", s, c)
        sum = (sum + combgeh(N, s)) --% MOD
    end
    return sum % MOD
end

local X, Y = getX(), getY()
local Total = modpow(3, N, MOD)
--dbgpr(Total, X, Y)
local ans = (Total - (3 * X) % MOD + (3 * Y) % MOD) % MOD
--local ans = (Total - (3 * X) + (3 * Y)) % MOD
print(ans)