Given are integers N and X. For each integer k between 0 and X (inclusive), find the answer to the following question, then compute the sum of all those answers, modulo 998244353.

* Let us repeat the following operation on the integer k. Operation: if the integer is currently odd, subtract 1 from it and divide it by 2; otherwise, divide it by 2 and add 2^{N-1} to it. How many operations need to be performed until k returns to its original value? (The answer is considered to be 0 if k never returns to its original value.)

Constraints

* 1 \leq N \leq 2\times 10^5
* 0 \leq X < 2^N
* X is given in binary and has exactly N digits. (In case X has less than N digits, it is given with leading zeroes.)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N
X


Output

Print the sum of the answers to the questions for the integers between 0 and X (inclusive), modulo 998244353.

Examples

Input

3
111


Output

40


Input

6
110101


Output

616


Input

30
001110011011011101010111011100


Output

549320998
def o_divisors(n):
    L = [i for i in range(1, int(n**0.5)+5) if n%i == 0]
    L += [n//i for i in L]
    L = sorted(list(set(L)))
    return [l for l in L if n//l % 2]
    
N = int(input())
T = input()
P = 998244353
bt = int(T, 2)
c = 0
ans = 0
L = []
for k in o_divisors(N):
    tt = T[:k]
    tt_ = tt.replace("0", "2").replace("1", "0").replace("2", "1")
    ttf = tt + (tt_ + tt) * ((N//k-1)//2)
    aa = int(tt, 2) + (int(ttf, 2) <= bt)
    for l, cc in L:
        if k % l == 0:
            aa -= cc
    L.append((k, aa))
    c += aa
    ans += k*2*aa
        
ans = (ans + 2 * N * (bt - c + 1)) % P
print(ans)
