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}.

Examples

Input

4
1231


Output

1


Input

10
2311312312


Output

0
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

N = int(readline())
A = list(map(int, read().decode().rstrip()))

U = N + 10
fact_ord = [0] * U
for n in range(1,U):
    fact_ord[n] = fact_ord[n//2] + (n//2)

def solve_01(A):
    ret = 0
    N = len(A) - 1
    comb_ord = [fact_ord[N] - fact_ord[i] - fact_ord[N-i] for i in range(N+1)]
    return sum(x * (y==0) for x,y in zip(A, comb_ord)) % 2

def solve(A):
    if len(A) == 1:
        return A[0]
    A = [abs(x - y) for x,y in zip(A, A[1:])]
    if all(x != 1 for x in A):
        B = [x // 2 for x in A]
        return 2 * solve_01(B)
    # 1 が存在
    A = [x % 2 for x in A]
    return solve_01(A)

print(solve(A))
