Problem:
Score : 700 points

Problem Statement

Given is a sequence of N digits a_1a_2\ldots a_N, where each element is 1, 2, or 3.
Let x_{i,j} defined as follows:

x_{1,j} := a_j \quad (1 \leq j \leq N)

x_{i,j} := | x_{i-1,j} - x_{i-1,j+1} | \quad (2 \leq i \leq N and 1 \leq j \leq N+1-i)

Find x_{N,1}.

Constraints

2 \leq N \leq 10^6

a_i = 1,2,3 (1 \leq i \leq N)

Input

Input is given from Standard Input in the following format:

N
a_1a_2\ldotsa_N

Output

Print x_{N,1}.

Sample Input 1

4
1231

Sample Output 1

1

x_{1,1},x_{1,2},x_{1,3},x_{1,4} are respectively 1,2,3,1.

x_{2,1},x_{2,2},x_{2,3} are respectively |1-2| = 1,|2-3| = 1,|3-1| = 2.

x_{3,1},x_{3,2} are respectively |1-1| = 0,|1-2| = 1.

Finally, x_{4,1} = |0-1| = 1, so the answer is 1.

Sample Input 2

10
2311312312

Sample Output 2

0

Solution in Julia (0.5.0) (judge verdict: Wrong Answer)

parseInt(x) = parse(Int, x)
parseMap(x::Array{SubString{String},1}) = map(parseInt, x)

function main()
	n = readline() |> parseInt
	a = parseMap(split(chomp(readline()),""))
	sa = sum(a)
	f = 0
	if n==2
		println(abs(a[1]-a[2]))
	else
		if (a[1]==1&&sa==3*(n-1)+1)||(a[n]==1&&sa==1+3*(n-1))||(a[1]==3&&sa==n+2)||(a[n]==3&&sa==n+2)
			println(2)
		else
			bb = Int[]
			b = zeros(Int,n-1)
			l = 1
			push!(bb,abs(a[1]-a[2]))
			b[1]=abs(a[1]-a[2])
			for i in 2:n-1
				v = abs(a[i]-a[i+1])
				b[i]=v
				if v!=0||bb[l]!=0
					push!(bb,v)
					l+=1
				end
			end
			if length(bb)==1
				println(bb[1])
			elseif length(bb)==2
				println(abs(bb[1]-bb[2]))
			else
				c = Int[]
				m = 1
				push!(c,abs(b[1]-b[2]))
				for i in 2:n-2
					v = abs(b[i]-b[i+1])
					if v!=0||c[m]!=0
						push!(c,v)
						m+=1
					end
				end
				if length(c)==1
					println(c[1])
				elseif length(c)==2
					println(abs(c[1]-c[2]))
				else
					v1 = zeros(Int,m)		
					v2 = zeros(Int,m)		
					v3 = ones(Int,m)		
					v4 = ones(Int,m)
					v1[1]=1
					v2[m]=1
					v3[1]=0
					v4[m]=0
					println(c==v1||c==v2||c==v3||c==v4?1:0)
				end
			end
		end
	end
end

main()