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 primeFactor(N):
    i, n, ret, d, sq = 2, N, {}, 2, 99
    while i <= sq:
        k = 0
        while n % i == 0: n, k, ret[i] = n//i, k+1, k+1
        if k > 0 or i == 97: sq = int(n**(1/2)+0.5)
        if i < 4: i = i * 2 - 1
        else: i, d = i+d, d^6
    if n > 1: ret[n] = 1
    return ret
def divisors(N):
    pf = primeFactor(N)
    ret = [1]
    for p in pf:
        ret_prev = ret
        ret = []
        for i in range(pf[p]+1):
            for r in ret_prev:
                ret.append(r * (p ** i))
    return sorted(ret)
def o_divisors(N):
    return [a for a in divisors(N) if N//a % 2 and a < N]
    
N = int(input())
T = input()
P = 998244353
ods = o_divisors(N)

bt = int(T, 2)
c = 0
ans = 0
L = []
for k in ods:
    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)
    if int(ttf, 2) <= bt:
        aa += 1
    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)

