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 DBG = false
local function dbgpr(...)
    if DBG then
        io.write("[dbg]")
        print(...)
    end
end

-- returns (b^n) % m
local function modpow(b, n, m)
    if n == 0 then
        return 1
    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 function fill90000(t, v)
    for i=1,900000 do
        t[i] = v
    end
end

local function default0(t)
    local mt = {
        __index = function(t,k)
            return 0
        end
    }
    setmetatable(t, mt)
end

-- ∑A = S/2 and ∑B = S/2 and ∑C = 0
local function getY()
    if S % 2 == 1 then
        return 0
    end
    local cache = {}
    default0(cache)
    --fill90000(cache, 0)
    cache[0] = 1
    local curmax = 0
    for i=1,N do
        local cur = a[i]
        curmax = curmax + cur
        for s=curmax,cur,-1 do
            cache[s] = (cache[s] + cache[s-cur]) % MOD
        end
    end
    return cache[math.floor(S/2)]
end

-- ∑A >= S/2
local function getX()
    local cache = {}
    default0(cache)
    --fill90000(cache, 0)
    cache[0] = 1
    local curmax = 0
    for i=1,N do
        local cur = a[i]
        curmax = curmax + cur
        for s=curmax,cur,-1 do
            cache[s] = (2 * cache[s] + cache[s-cur]) % MOD
        end
        for s=cur-1,0,-1 do
            cache[s] = 2 * cache[s] % MOD
        end
    end
    local sum = 0
    for s = math.ceil(S/2), curmax, 1 do
        sum = sum + cache[s]
    end
    return sum
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
print(ans)