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
MOD = 998244353

def getDivisors(x):
    anss = set()
    for d in range(1, int(x**0.5)+1):
        if x%d == 0:
            anss.add(d)
            anss.add(x//d)
    return sorted(anss)

N = int(input())
X = input()

divisors = getDivisors(2*N)

periods = [d for d in divisors if d%2 == 0 and (2*N//d)%2]
numP = len(periods)

anss = dict(zip(periods, [0]*numP))
for p in periods:
    lenPtn = p//2
    anss[p] += int(X[:lenPtn], 2) % MOD
    Ts = X[:lenPtn]
    notTs = ''.join(['1' if T == '0' else '0' for T in Ts])
    T2s = (Ts+notTs)*(N//p) + Ts
    if T2s <= X:
        anss[p] += 1
    anss[p] %= MOD
    for p2 in periods:
        if p == p2: continue
        if p2%p == 0:
            anss[p2] -= anss[p]

ans = sum([k*v for k, v in anss.items()])
print(ans % MOD)
